Codeforces Round #239 (Div. 2) C Triangle(暴力)

本文探讨了如何确定一个特定的直角三角形是否能够放置于坐标平面中,使其边不平行于坐标轴,并且所有顶点都有整数坐标。通过编程的方法给出了具体的解决方案和示例。

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思路:固定一个点(0,0)然后暴力枚举计算就好了,注意a和b的夹角不一定是90°的


#include<bits/stdc++.h>
using namespace std;
#define eps 1e-7
int main()
{
    int a,b,i,j,i1,j1;
    scanf("%d%d",&a,&b);
    int ok=0;
    int flag=0;
    for(i=1;i<a;i++)
    {
        for(j=1;j<b;j++)
		{
            i1=sqrt(a*a-i*i);
            j1=sqrt(b*b-j*j); 
            if(i1==sqrt(a*a-i*i)&&j1==sqrt(b*b-j*j))
            {
                if(i!=j1&&i*j1==i1*j)
                {
                    ok=1;
                    flag=1;
                    break;
                }
            }
        }
        if(flag)  break;
    }
    if(!ok)  
		printf("NO\n");
    else
    {
		printf("YES\n");
		printf("0 0\n");
		printf("%d %d\n",-i1,i);
		printf("%d %d\n",j,j1);
    }
}


C. Triangle
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.

Input

The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.

Output

In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.

Examples
input
1 1
output
NO
input
5 5
output
YES
2 1
5 5
-2 4
input
5 10
output
YES
-10 4
-2 -2
1 2

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