the Sum of Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 162 Accepted Submission(s): 101
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].
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.
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
1 3
2 5
Sample Output
Case #1: 36
Case #2: 224
Case #2: 224
Source
1^3+2^3+4^3.....+n^3=(n*(n+1)/2)^2
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std; int main()
{
int cas;
__int64 a,b;
scanf("%d",&cas);
for(int i=;i<=cas;i++)
{
scanf("%I64d%I64d",&a,&b);
printf("Case #%d: %I64d\n",i,(b*(b+)/)*(b*(b+)/)-(a*(a-)/)*(a*(a-)/));
}
return ;
}