hdu 5053 the Sum of Cube(上海网络赛）

最新推荐文章于 2021-10-18 14:44:23 发布

转载最新推荐文章于 2021-10-18 14:44:23 发布 · 67 阅读

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原文链接：http://www.cnblogs.com/mfrbuaa/p/4292719.html

文章标签：

#java

本文介绍了如何通过算法计算指定范围内的整数立方和，并提供了优化的代码实现。

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

    2 1 3 2 5
   

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;
}

转载于:https://www.cnblogs.com/mfrbuaa/p/4292719.html