the Sum of Cube 就是立方和

最新推荐文章于 2018-07-24 09:38:43 发布

原创最新推荐文章于 2018-07-24 09:38:43 发布 · 622 阅读

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本文介绍了一个关于计算指定范围内整数立方和的问题及其解决方案。通过输入一对整数定义的范围[A,B]，程序将计算并输出该范围内所有整数的立方和。示例展示了具体的输入输出情况，并提供了一段C语言代码实现。

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the Sum of Cube

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 84 Accepted Submission(s) : 31

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

就是简单的[a,b]的立方根，注意范围

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main() {
    long long t,t1,a,b,s,i,sum;
    scanf("%lld",&t1);
    for(t=1; t<=t1; t++) {
        sum=0;
        scanf("%lld%lld",&a,&b);
        for(i=a; i<=b; i++) {
            sum=sum+i*i*i;
        }
        printf("Case #%lld: %lld\n",t,sum); 
    }

    return 0;
}
//Case #1: 36
//shyazhut 13:26:53