the Sum of Cube hd 5053

q923714892發表於2020-04-06
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<stdio.h>
int main()
{
	int i,a,b,n;
	int cas=0;
	__int64 sum;
	scanf("%d",&n);
	while(n--)
	{
		cas++;
		sum=0;
		scanf("%d%d",&a,&b);
		for(i=a;i<=b;i++)
		{
			sum+=(__int64)i*i*i;
		}
		printf("Case #%d: %I64d\n",cas,sum);
	}
	return 0;
}


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