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

2
1 2
2 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;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4237316.html