poj2924---高斯求和

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n,tim=0;
    long long ans,a,b;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lld %lld",&a,&b);
        ans = (a+b)*(b-a+1)/2;
        printf("Scenario #%d:
",++tim);
        printf("%lld

",ans);
    }
}
View Code

long long 的范围:9223372036854775807~-9223372036854775808  亿万亿

原文地址:https://www.cnblogs.com/gabygoole/p/4588263.html