hdu 5053 the Sum of Cube(上海网络赛）

最新推荐文章于 2017-08-06 19:20:08 发布

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本文介绍了一个算法问题，即计算给定整数区间内所有整数的立方和。输入包括多个测试用例，每个用例由一对整数表示区间范围，输出为对应区间内所有整数立方和的结果。

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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): 405 Accepted Submission(s): 224

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

Sample Output

  
   Case #1: 36
Case #2: 224

Source

2014 ACM/ICPC Asia Regional Shanghai Online

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
    long long a,b;
    int ca=1;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&a,&b);
        printf("Case #%d: %I64d\n",ca++,(b*(b+1)/2)*(b*(b+1)/2)-(a*(a-1)/2)*(a*(a-1)/2));
    }
    return 0;
}