hdu 5053 the Sum of Cube

**Problem 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
2
1 3
2 5

Sample Output
Case #1: 36
Case #2: 224
**
水题，没什么思路好说的。
翻译：t组样例，求出在a到b范围内所有数的立方和，按照题目格式输出
唯一坑的地方就是注意数据范围，记得用long long

#include<stdio.h>#include<math.h>#include<stdlib.h>int main(){    int t,a,b,i,j;    long long sum;    scanf("%d",&t);    for( i = 1; i <= t; i ++)    {        scanf("%d%d",&a,&b);        sum = 0;        for( j = a; j <= b; j ++)        {            sum += pow(j,3);        }        printf("Case #%d: %lld\n",i,sum);    }    return 0;}