hdoj5053the Sum of Cube

最新推荐文章于 2017-08-06 19:20:08 发布

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本文介绍了一种通过编程实现的算法，该算法可以计算给定整数范围内的所有整数立方值之和。输入包括一系列测试用例，每个用例包含一对整数，分别表示区间的开始和结束。输出则是每个测试用例中指定区间内所有整数立方值的总和。

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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

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    double sum,i;//考虑到大数，sum和i定义为double型.
    int t,n,m,l=0;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        
        scanf("%d%d",&n,&m);
        for(i=n;i<=m;i++)
        {
            sum+=i*i*i;
        }
        l++;
        printf("Case #%d: %.lf\n",l,sum);
    }
}