（一）：the Sum of Cube(立方和。。。。高效率）

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Description

A range is given, the begin and the end are both integers. You should sum the cube of all the integers in the range.

Input

The first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve.
Each case of input is a pair of integer A,B(0 < A <= B <= 10000),representing the range[A,B].

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then output the answer – sum the cube of all the integers in the range.

Sample Input

21 32 5

Sample Output

Case #1: 36Case #2: 224

看表格求立方和

4444444444444444444444444444444444444444333333444433333344443333334444222333444422233344441223334444 看表后，很容易看出来求1到4的立方和就是求这个（单元格边长为1）正方形表格的面积；；；（假设求从1到n的之间的数的立方和）则表格边长为n*(n+1)/2...代码如下：

#include<stdio.h>int main(){int t;double m,n,sum;double s1,s2;int cont=0;scanf("%d",&t);while(t--){scanf("%lf %lf",&m,&n);     //the range[m,n]s1=(m*(m-1)/2)*(m*(m-1)/2);   //从1到m-1的数的立方和 s2=(n*(n+1)/2)*(n*(n+1)/2);   //从1到n的数的立方和 sum=s2-s1;//从m到n的数的立方和printf("Case #%d: %.lf\n",++cont,sum); }return 0;}

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