Đưa ra một số N, nhiệm vụ là tìm tổng yếu tố lẻ. Ví dụ: Show Input : n = 30 Output : 24 Odd dividers sum 1 + 3 + 5 + 15 = 24 Input : 18 Output : 13 Odd dividers sum 1 + 3 + 9 = 13 Đặt P1, P2, PK PK là yếu tố chính của n. Đặt A1, A2, .. Ak là sức mạnh cao nhất của P1, P2, .. PK tương ứng chia n, tức là, chúng ta có thể viết n là n = (p1a1)* (p2a2)*, (pkak).n = (p1a1)*(p2a2)* … (pkak). Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak) Để tìm ra tổng các yếu tố lẻ, chúng ta chỉ cần bỏ qua các yếu tố và sức mạnh của chúng. Ví dụ: xem xét n = 18. Nó có thể được viết là 2132 và tổng của tất cả các yếu tố là (1)*(1 + 2)*(1 + 3 + 32). Tổng các yếu tố lẻ (1)*(1+3+32) = 13. Để loại bỏ tất cả các yếu tố chẵn, chúng tôi liên tục chia n trong khi nó chia hết cho 2. Sau bước này, chúng tôi chỉ nhận được các yếu tố lẻ. Lưu ý rằng 2 là thậm chí còn duy nhất. python3# Chương trình Python3 dựa trên công thức# để tìm tổng của tất cả các ước số# của N.Import Math # Trả về tổng của tất cả các yếu tố# của n.def sumofoddfactors (n): & nbsp; & nbsp; & nbsp; & nbsp;# đi qua tất cả & nbsp; & nbsp;# các yếu tố chính. & nbsp; & nbsp; res = 1 & nbsp; & nbsp; & nbsp; & nbsp;# bỏ qua các yếu tố thậm chí bởi & nbsp; & nbsp;# của 2 & nbsp; & nbsp; trong khi n % 2 == 0: & nbsp; & nbsp; & nbsp; & nbsp; n = n // 2 & nbsp; & nbsp; & nbsp; & nbsp; cho i trong phạm vi (3, int (math.sqrt (n) + 1)): & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;# Trong khi tôi chia n, in & nbsp; & nbsp; & nbsp; & nbsp;# i và chia n & nbsp; & nbsp; & nbsp; & nbsp; đếm = 0 & nbsp; & nbsp; & nbsp; & nbsp; curr_sum = 1 & nbsp; & nbsp; & nbsp; & nbsp; curr_term = 1 & nbsp; & nbsp; & nbsp; & nbsp; trong khi n % i == 0: & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; đếm+= 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; n = n // i & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; curr_term *= i & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; curr_sum += curr_term & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; res *= curr_sum & nbsp; & nbsp; & nbsp; & nbsp;# Điều kiện này là & nbsp; & nbsp;# xử lý trường hợp khi & nbsp; & nbsp;# n là một số nguyên tố. & nbsp; & nbsp; nếu n> = 2: & nbsp; & nbsp; & nbsp; & nbsp; res *= (1 + n) & nbsp; & nbsp; & nbsp; & nbsp; return res # Trình điều khiển CODEN = 30print (sumofoddfactors (n)) # Mã này được đóng góp bởi trên Shar Sharad_bhardwaj. Output: 24 Độ phức tạp về thời gian: O (SQRT (N)): O(sqrt(n)) Không gian phụ trợ: O (1): O(1) Vui lòng tham khảo hoàn thành bài viết về tìm tổng các yếu tố kỳ lạ của một số để biết thêm chi tiết! Cải thiện bài viết Lưu bài viết Cải thiện bài viết Lưu bài viết ĐọcN, the task is to check if N has an odd number of odd divisors and even number of even divisors. Examples::
Đầu vào: n = 36Output: & nbsp; yesexplanation: các ước số của 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36count của các ước số lẻ (1, 3, 9) = 3 [LOD] (2, 4, 6, 12, 18, 36) = 6 [chẵn]: The idea is to find the factors of the number N and count the odd factors of N and even factors of N. Finally, check if the count of odd factors is odd and count of even factors is even. Đầu vào: & nbsp; n & nbsp; = & nbsp; 28Output: & nbsp; không C++
Cách tiếp cận ngây thơ: Ý tưởng là tìm ra các yếu tố của số N và đếm các yếu tố kỳ lạ của N và thậm chí các yếu tố của N. Cuối cùng, hãy kiểm tra xem số lượng các yếu tố lẻ là số lẻ và số lượng của các yếu tố thậm chí là chẵn.
Dưới đây là việc thực hiện phương pháp trên: Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)2
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)6 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)7 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)9
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)5 240 241 242 249 #include <bits/stdc++.h> 0243 241 245 249 #include <bits/stdc++.h> 4243 #include <bits/stdc++.h> 6246 241 248 240 241 242 249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4243 241 245 249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4243 #include <bits/stdc++.h> 6240 #include <bits/stdc++.h> 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 6246 241 248 240 std; 8246 #include <bits/stdc++.h> 2Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2243 #include <bits/stdc++.h> 2 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0
246 241 namespace 1Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 4Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 std; 6
240#define lli long long int0#define lli long long int1 #define lli long long int2240 #define lli long long int 0#define lli long long int 7 #define lli long long int 2
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 8 void 9Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Java
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0
240 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)03 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)04 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)05 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)044 243 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 checkFactors(lli N) 8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)12 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 249 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)41 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 246 #include <bits/stdc++.h> 2249 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)47 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 243 #include <bits/stdc++.h> 6243 #include <bits/stdc++.h> 2243 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 checkFactors(lli N) 8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)12 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 249 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)41 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 243 241 245 249 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)47 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 246 241 248 249 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)41 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 246 #include <bits/stdc++.h> 2249 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)47 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 243 #include <bits/stdc++.h> 6240 #include <bits/stdc++.h> 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 6246 #include <bits/stdc++.h> 2243 #include <bits/stdc++.h> 2 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2246 241 namespace 1
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 std; 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 240 #define lli long long int 0#define lli long long int 1 #define lli long long int 2Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6
Python3240 #define lli long long int 0#define lli long long int 7 #define lli long long int 2
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 8 void 9Java
246 #include <bits/stdc++.h> 22499 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)03 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)04 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)05 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)044 243 #include <bits/stdc++.h> 22499
246 #include <bits/stdc++.h> 22499 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)03 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)04 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)05 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)044 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 checkFactors(lli N) 8 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)12 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 249 2439 2462 2440 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 246 #include <bits/stdc++.h> 22499 249 #include <bits/stdc++.h> 0124624____240 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)13 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 #include <bits/stdc++.h> 612468 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)36 2440 2440 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)04 #include <bits/stdc++.h> 67240 #include <bits/stdc++.h> 012468 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)36 2440 2440 240 #include <bits/stdc++.h> 777____110#define lli long long int 1 2462 #include <bits/stdc++.h> 81Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 22499 240 #include <bits/stdc++.h> 777____110#define lli long long int 7 2462 #include <bits/stdc++.h> 81241 #include <bits/stdc++.h> 922440 2440 #include <bits/stdc++.h> 952499 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 2479 2440 2430 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6C#
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 using 16Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 checkFactors(lli N) 8 using 21Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)15 using 23Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 240 241 using 28240 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 243 241 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)30 243 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 246 241 248 249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4243 #include <bits/stdc++.h> 6243 #include <bits/stdc++.h> 2243 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 246 241 248 249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4246 241 namespace 1249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4243 #include <bits/stdc++.h> 6240 #include <bits/stdc++.h> 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 using 77240 using 79#define lli long long int 1 2410 2411 2412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2240 using 79#define lli long long int 7 2410 2411 2412
2422 checkFactors(lli N) 5 void using 96Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 namespace 00Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6
JavaScript
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 namespace 10Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 namespace 13240 namespace 15240 241 242 243 241 namespace 21246 241 248 249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4243 #include <bits/stdc++.h> 6249 #include <bits/stdc++.h> 0246 241 248 249 #include <bits/stdc++.h> 0249 #include <bits/stdc++.h> 0249 #include <bits/stdc++.h> 4246 241 namespace 1249 #include <bits/stdc++.h> 0246 #include <bits/stdc++.h> 2249 #include <bits/stdc++.h> 4243 #include <bits/stdc++.h> 6240 #include <bits/stdc++.h> 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 6Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 using 77240 std; 8240 using 79#define lli long long int 1 2410 2411 2412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2240 using 79#define lli long long int 7 2410 2411 2412
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 namespace 81Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6
JavaScript C++
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 namespace 13Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 240 241 242 243 241 namespace 21
243 #include <bits/stdc++.h> 2 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 246 #include <bits/stdc++.h> 2246 241 namespace 47Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 std; 6
240 namespace 66#define lli long long int 1 2410 namespace 692412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 4Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6240 namespace 66#define lli long long int 7 2410 namespace 692412
Độ phức tạp về thời gian: O (N (1/2))
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 using 16Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)4 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 checkFactors(lli N) 8 using 21
240 241 using 28Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 243 241 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)30 240 2408 #define lli long long int 12412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2240 2408 #define lli long long int 72412
246 241 248 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 249 #include <bits/stdc++.h> 0Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6
Python3246 241 namespace 1Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 using 77240 using 79#define lli long long int 1 2410 2411 2412 240 using 79#define lli long long int 7 2410 2411 2412 2422 checkFactors(lli N) 5 void using 96Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 checkFactors(lli N) 8 namespace 00240 #include <bits/stdc++.h> 77Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 #define lli long long int 12412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 22499 240 #include <bits/stdc++.h> 77Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)10 #define lli long long int 72412 241 #include <bits/stdc++.h> 922440 2440 #include <bits/stdc++.h> 952499 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 2479 2440 2430 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)07 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6C#
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 namespace 93 #define lli long long int 52Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 8 #define lli long long int 55
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 std; 61240 using 79#define lli long long int 12412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2240 using 79#define lli long long int 72412
2422 checkFactors(lli N) 5 void using 96Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 0 namespace 00Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6
JavaScript
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 std; 46Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 8 #define lli long long int 97
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)0 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 241 std; 13240 namespace 66#define lli long long int 1 2410 namespace 692412 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 #include <bits/stdc++.h> 2240 namespace 66#define lli long long int 7 2410 namespace 692412
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 namespace 00Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)1 void 6
Độ phức tạp về thời gian: O (log (n)) vì nó đang sử dụng hàm sqrt sẵn cóO(log(N)) because it is using inbuilt sqrt function Không gian phụ trợ: O (1)O(1) |