codeforces 407C Triangle

最新推荐文章于 2020-08-15 21:48:02 发布

原创最新推荐文章于 2020-08-15 21:48:02 发布 · 339 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

codeforce 专栏收录该内容

32 篇文章

订阅专栏

本文探讨了如何确定一个直角三角形是否可以放置在二维平面上，使其边不平行于坐标轴，并提供了具体的实现算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description
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 Input
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：给出一个直角三角形的两条边长a、b，看能否将这个三角形放到一个二维平面内，使得三条边都不能平行于坐标轴，如果能输出三个点的坐标，否则输出NO

把直角顶点放在（0，0）会更好做；首先，看两个边长是不是都能成为两个整数的平方和的开根号（可能会有多个！！），然后判断以这两个边为斜边的直角三角形是不是能想似，最后就看直角三角形最后一条边是否是平行于坐标轴的。

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int a,b,h1[10],l1[10],h2[10],l2[10],flag1=0,flag2=0;
    cin>>a>>b;
    for(int i=1; i<1000; i++)
    {
        for(int j=i; j<1000; j++)
        {
            if(i*i+j*j==a*a)
            {
                h1[flag1]=i;
                l1[flag1]=j;
                flag1++;
            }
            if(i*i+j*j==b*b)
            {
                h2[flag2]=i;
                l2[flag2]=j;
                flag2++;
            }
        }
    }
    int t=0;
    if(flag1==0||flag2==0)
        cout<<"NO"<<endl;
    else
    {
        for(int i=0; i<flag1; i++)
        {
            for(int j=0; j<flag2; j++)
            {
                if(1.0*h1[i]/h2[j]==1.0*l1[i]/l2[j])
                {
                    if(h1[i]!=l2[j])
                    {
                        cout<<"YES"<<endl;
                        cout<<0<<" "<<0<<endl;
                        cout<<l1[i]<<" "<<h1[i]<<endl;
                        cout<<0-h2[j]<<" "<<l2[j]<<endl;
                        t=1;
                        break;
                    }
                    else if(h2[j]!=l1[i])
                    {
                        cout<<"YES"<<endl;
                        cout<<0<<" "<<0<<endl;
                        cout<<l2[j]<<" "<<h2[j]<<endl;
                        cout<<0-h1[i]<<" "<<l1[i]<<endl;
                        t=1;
                        break;
                    }
                }
            }
            if(t==1)
                break;
        }
        if(t==0)
            cout<<"NO"<<endl;
    }
    return 0;
}