【HDU

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the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3082    Accepted Submission(s): 1295

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

21 32 5

 

Sample Output

Case #1: 36Case #2: 224

//题意是求A—B各个立方数之和  比如1,3 答案=1+2*2*2+3*3*3=36。

// 要注意的是_int64的 scanf printf 用法   // 相当于long long int

//代码如下：

#include <stdio.h>int main(){__int64  a,b,sum=0,i,k;int t;scanf("%d",&t);k=1;while(t--){sum=0;scanf("%I64d%I64d",&a,&b);for(i=a;i<=b;i++) sum+=i*i*i; printf("Case #%d: %I64d\n",k,sum);k++;}return 0;}