An Easy Problem

描述

    In this problem, you are given two integers L and R, and your task is to calculate the sum of all the number's square between L and R(inclusive).

输入

The first line contains an integer T, indicates the number of test case.
The next T lines, each line contains two integers L and R(1≤L,R≤10^5).

输出

Print an integer represents the sum.

样例输入

21 22 4

样例输出

5

29

//代码如下

#include<iostream>#include<stdio.h>using namespace std;int main(){int n;double a,b,i,sum;while(cin>>n){while(n--){cin>>a>>b;sum=0;if(a>b) swap(a,b);for(i=a;i<=b;i++)sum+=i*i;printf("%.lf\n",sum);//此处不能用cout，因为数较大，不能完整的输出}}return 0;}

//参考别人的代码（用的是long long型的）

#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;inline LL get_sum(LL x) {    return x * (x + 1) * (2 * x + 1) / 6;//从1到x的平方的和的公式}int main() {    int T;    LL L, R;    scanf("%d", &T);    while(T--) {        scanf("%lld%lld",&L, &R);        if(L > R) swap(L, R);        printf("%lld\n", get_sum(R) - get_sum(L - 1));    }    return 0;}