ACM Problem Description求11+22+..+n*n ,n不超过2500Input输入由多行组成,每行一个整数nOutput每行输入对应一行输出,为所求结果Sample Input12Sample Output15Author陈燕晖#includeint main(){ int n,s;\x05while(scanf("%d",&n)

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ACM Problem Description求1*1+2*2+..+n*n ,n不超过2500Input输入由多行组成,每行一个整数nOutput每行输入对应一行输出,为所求结果Sample Input12Sample Output15Author陈燕晖#includeint main(){ int n,s;\x05while(scanf("%d",&n)
ACM
Problem Description
求1*1+2*2+..+n*n ,n不超过2500
Input
输入由多行组成,每行一个整数n
Output
每行输入对应一行输出,为所求结果
Sample Input
1
2
Sample Output
1
5
Author
陈燕晖
#include
int main()
{ int n,s;
\x05while(scanf("%d",&n)!=EOF)
\x05{ s=0;
\x05\x05if(n

不知道你们的OJ系统是支持__int64还是long long,因为n的三次方已经超过了int所能表示的范围,所以我们要把结果定义成__int64或者long long型的.代码如下：

#include<stdio.h>

int main()
{
        __int64 n,s;
        while(scanf("%I64d",&n)!=EOF)
{
        s=0;
if(n<=2500)
{
    s=(n*(n+1)*(2*n+1))/6;
            printf("%I64d\n",s);
}

}
return 0;
}

或者：

#include<stdio.h>

int main()
{
        long long n,s;
        while(scanf("%Ild",&n)!=EOF)
{
        s=0;
if(n<=2500)
{
    s=(n*(n+1)*(2*n+1))/6;
            printf("%lld\n",s);
}

}
return 0;
}

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