Codeforces Round #239 (Div. 2) 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.

Sample test(s)

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

题意 ：就是告诉你一个直角三角形的两条直角边的长度，是否能找出这个三角形的三个点的坐标都是整数，并且任何一条边都不能与坐标轴平行，如果不满足输出NO

思路 ： 这个题算是从头开始枚举，以原点为基准开始转，所以原点算一个点，然后从1开始枚举到a，因为不能与坐标轴平行的话，其中一个点到原点的那条边不能与坐标轴平行，所以只能作为坐标轴中的一条斜边来算（不是三角形的斜边），所以肯定比a小，然后枚举出整数来之后，再去求剩下的那个点就行了

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

int main()
{
    int a,b;
    scanf("%d %d",&a,&b) ;
    for(int i =  ; i < a ; i++)
    {
        double t = sqrt(a*a-i*i);
        if(t-(int)t==)
        {
            double ta = -*(b*t)/a;
            double tb = (b*i)/a;
            if(ta-(int)ta== && tb-(int)tb== && tb!=t)
            {
                printf("YES\n0 0\n%d %.0lf\n%.0lf %.0lf\n",i,t,ta,tb) ;
                return ;
            }
        }
    }
   printf("NO\n") ;
    return ;
}