【HD 5053】the Sum of Cube

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

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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;
}