杭电 5053 the Sum of Cube（求区间内的立方和）打表法

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

 

#include<cstdio>
__int64 a[1000000]={0,1};
int main()
{
    for(__int64 i=1;i<=1000000;i++)
    {
        a[i]=a[i-1]+i*i*i;
    }
    __int64 t,k=0;;
    scanf("%I64d",&t);
    while(t--)
    {
        __int64 m,n;
        scanf("%I64d%I64d",&m,&n);
        printf("Case #%I64d: %I64d
",++k,a[n]-a[m-1]);
    }
}