【HD 5053】the Sum of Cube

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本文介绍了一种通过预计算方式快速解决区间内整数立方和问题的方法。利用C++编程语言实现了一个简单高效的程序，该程序可以处理多个测试用例，并在给定范围内计算所有整数的立方和。

the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2459 Accepted Submission(s): 1032

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

——————————————————————————————
超级水的一道题
本来要用下面的方法图解的，但是直接暴力出来了……

4   4   4   4   4   4   4    4    4    4     
4   4   4   4   4   4   4    4    4    4  
4   4   4   4   4   4   4    4    4    4   
4   4   4   4   4   4   4    4    4    4           
3   3   3   3   3   3   4    4    4    4 
3   3   3   3   3   3   4    4    4    4 
3   3   3   3   3   3   4    4    4    4 
2   2   2   3   3   3   4    4    4    4 
2   2   2   3   3   3   4    4    4    4 
1   2   2   3   3   3   4    4    4    4

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
__int64 a[10005];
void vo()
{
    __int64 i;
    for( i=1; i<10005; i++)
    {
        a[i]=i*i*i;
    }
}
int main()
{
    vo();
    __int64 j,sum;
    int t,x,y,k=1;
    scanf("%d",&t);
    while( t-- )
    {
        sum=0;
        scanf("%d%d",&x,&y);
        for( j=x; j<=y; j++ )
        {
            sum=a[j]+sum;
        }
        printf("Case #%d: %I64d\n",k,sum);
        k++;
    }
    return 0;
}