hdoj Calculate S(n) 2114 （数学规律 取余）

Calculate S(n)

Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9401 Accepted Submission(s): 3415

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

12

Sample Output

00010009
//S（n）=1^3+2^3+3^3+......+n^3可转化为S(n)=（1+2+3+......+n）^2;
知道这个转化就简单了。
#include<stdio.h>#include<string.h>#include<algorithm>#define ll long longusing namespace std;int main(){ll n;while(scanf("%lld",&n)!=EOF){ll sum=((n*(n+1)/2%10000)*(n*(n+1)/2%10000))%10000;printf("%04lld\n",sum);}return 0;}