hdu 2132 An easy problem

#include<stdio.h>
long long sum[100010];
int main()
{
    int n;
    int i;

    sum[0]=0,sum[1]=1;
    for(i=2; i<100010; i++)
    {
        if(i%3==0)sum[i]=sum[i-1]+(long long)i*i*i;
        else sum[i]=sum[i-1]+i;
    }

    while(scanf("%d",&n)!=EOF&&n>=0)
    {
        printf("%I64d\n",sum[n]);
    }
    return 0;
}

****需注意的地方：

if(i%3==0)sum[i]=sum[i-1]+(long long)i*i*i;

转载于:https://www.cnblogs.com/XDJjy/p/3180853.html