Calculate S(n) (规律题)

Calculate S(n)

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10508    Accepted Submission(s): 3771

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

 

题意：

给你一个n，问你1^3+2^3+3^3+4^3+...+n^3的后四位是多少。

解析：

就一个公式，（1^3+2^3+3^3+4^3+...+n^3） = （1 + 2 + 3 + 4 + ... + n）^ 2

注意取模就好了。

#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int mod = 1e4;
const double esp = 1e-6;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int maxn = 500005;
using namespace std;

int main(){
    LL n;
    while(cin>>n){
        LL num = (n * (n + 1)) >> 1;
        num %= mod;
        num *= num;
        printf("%04d\n",num%mod);
    }
    return 0;
}