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
做这道题只要知道立方和公式就非常简单了。
n的平方的前n项和公式:n(n+1)(2n+1)/6
ac代码:
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<cstring>
#include<string.h>
#include<queue>
using namespace std;
typedef long long ll;
long long a[50050];
int main()
{
ll n;
while(scanf("%lld",&n)!=EOF)//开始外面n设置成ll,里面结果用Int取得,wr了
{
printf("%04lld\n",(((n*(n+1)/2)%10000)*((n*(n+1)/2)%10000))%10000);这个地方要对10000取模
}
return 0;
}