| Marshal‘s confusion II | ||||||
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| Description | ||||||
| one day, Marshal want to show the answer :Calculate S(n). S(n)=1^3+2^3 +3^3 +......+n^3 . | ||||||
| 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
#include<cmath> #include<queue> #include<iostream> using namespace std; const int m=10000; typedef long LL; int main() { LL n; while(~scanf("%ld",&n))//1^3+2^3+...+n^3=[n(n+1)/2]^2 { LL a; a=n*(n+1)/2%m; LL sum=a*a%m; printf("%04d\n",sum); } return 0; }
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原文：http://www.cnblogs.com/beige1315402725/p/4993354.html
