Hdu 4486 Pen Counts

Pen Counts

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 130 Accepted Submission(s): 66

Problem Description

Chicken farmer Xiaoyan is getting three new chickens, Lucy, Charlie and CC. She wants to build a chicken pen so that each chicken has its own, unobstructed view of the countryside. The pen will have three straight sides; this will give each chicken its own side so it can pace back and forth without interfering with the other chickens. Xiaoyan finds a roll of chicken wire (fencing) in the barn that is exactly N feet long. She wants to figure out how many different ways she can make a three sided chicken pen such that each side is an integral number of feet, and she uses the entire roll of fence.
Different rotations of the same pen are the same, however, reflections of a pen may be different (see below).

Input

The first line of input contains a single integer P,(1<= P <=1000), which is the number of data sets that follow. Each data set should be processed identically and independently.

Each data set consists of a single line of input. It contains the data set number, K, and the length of the roll of fence, N, (3 <= N <= 10000).

Output

For each data set there is a single line of output. It contains the data set number, K, followed by a single space which is then followed by an integer which is the total number of different three-sided chicken pen configurations that can be made using the entire roll of fence.

Sample Input

5 1 3 2 11 3 12 4 100 5 9999

Sample Output

1 1 2 5 3 4 4 392 5 4165834

Source

Greater New York 2012 

已知三角形三边之和，求三角形的个数方法：枚举一条边（一般是最短），求令一条边的合法取值范围

时间复杂度O（n）

#include

using namespace std;

#define Min(a,b) (ab?a:b)

int main()
{
	int T,t;
	int a,s,min,max,L;
	int d;

	scanf("%d",&T);

	while(T--)
	{
		scanf("%d%d",&t,&s);

		d=0;
		for(a=1;a<=s/3;a++)
		{
			min=Max(a,s/2+1-a);		max=(s-a)/2;
			
			if(min==a)
			{
				d++;
				min++;
			}
			if(max=min)
				d+=(max-min+1)*2;
		}

		printf("%d %d\n",t,d);

	}

	return 0;
}