Triangle - CodeForces 407A 水题

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本文探讨了如何确定一个直角三角形是否能在平面上找到一个位置，使其边不平行于坐标轴，并且所有顶点都有整数坐标。提供了一个算法实现及样例输入输出。

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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

思路：枚举一条边的横坐标，注意第三条边别和坐标轴并行就行。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
void solve()
{int T,t,n,m,i,j,k,a,b,a1,a2=1,b1,b2;
  scanf("%d%d",&a,&b);
  for(a1=1;a1<a;a1++)
  { a2=sqrt(a*a-a1*a1);
    if(a*a==a1*a1+a2*a2)
    { b1=a1*b/a;
      b2=a2*b/a;
      if(b*b==b1*b1+b2*b2 && a2!=b1)
      { printf("YES\n");
        printf("0 0\n");
        printf("%d %d\n",a1,a2);
        printf("%d %d\n",-b2,b1);
        return;
      }
    }
  }
  printf("NO\n");
}
int main()
{ solve();
}