|
Calculate S(n) Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9640 Accepted Submission(s): 3492
Problem Description Calculate S(n).
S(n)=13+23 +33 +......+n3 .
Input Each line will contain one integer N(1 < n < 1000000000). Process to end of file.
Output For each case, output the last four dights of S(N) in one line.
Sample Input 1 2
Sample Output 0001 0009
Author 天邪
Source |
1 //1~N的立方和公式为S(N) = 1^3 + 2^3 + 3^3 + … + N^3 = N^2*(N+1)^2/4 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 int main() 7 { 8 long long n; 9 long long sum; 10 while(cin>>n){ 11 n%=10000; 12 sum=(n*n)*(n+1)*(n+1)/4; 13 sum%=10000; 14 printf("%04lld\n",sum); 15 } 16 return 0; 17 }
原文：http://www.cnblogs.com/shenyw/p/5176500.html