hdu 4486 Pen Counts

Pen Counts

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 246    Accepted Submission(s): 144

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

 

Recommend

liuyiding

 

这题考的是三角形的一些知识。

题意：

给你三角形的周长。问你能构成多少边为整数的三角形。如果一个三角形能通过另一个三角形绕某一顶点得到的话那么这两个三角形算一个。

思路：

先总结一下三角形的相关知识。

设三角形周长为len。三边关系为l3>=l2>=l1。

两边之和>第三边。

所以三角形中最长边的取值范围为

                len/3向上取整<=l3

rest=len-l3。那么第二大边取值范围为

                rest/2向下取整

对于一个等腰三角形其底边长的最大值<(len+1)/2。所以给定一个周长可以确定有多少个等腰或等边三角形。

所以每确定一个最长边那么三角形的个数是能确定的。

详细见代码：

#include 
#include
using namespace std;

int main()
{
    int len,rest,l3,durl2,p,cas,ans;

    scanf("%d",&p);
    while(p--)
    {
        scanf("%d%d",&cas,&len);
        ans=0;
        for(l3=(len+2)/3;l3<(len+1)/2;l3++)//枚举最长边的长度
        {
            rest=len-l3;//剩下的的长度
            durl2=l3-(rest-1)/2;//第二大边的取值区间
            ans+=2*durl2;//三边都不同算两个
            if(!(rest&1))//可组成等腰三角形
                ans-=1;
            ans-=1;
        }
        if(len%3==0)//可组成等边
            ans+=1;
        printf("%d %d\n",cas,ans);
    }
    return 0;
}