【杭电oj】5053-the Sum of Cube（立方和）

the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1984    Accepted Submission(s): 811

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

图形法推倒1的立方到n的立方求和的公式。推理过程如下：

计算结果如下：   1³+2³+3³+…+n³=（n*(n+1)/2）²

有公式代码就不是问题了。

#include <stdio.h>
int main()
{
	int u;
	int ans=1;
	double a,b;
	double c,c1,c2;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%lf %lf",&a,&b);
		printf ("Case #%d: ",ans++);
		c1=(a*(a-1)/2)*(a*(a-1)/2);
		c2=(b*(b+1)/2)*(b*(b+1)/2);
		c=c2-c1;
		printf ("%.lf\n",c);
	}
	return 0;
}