• CodeForces


    D. Vasya and Triangle
    time limit per test1 second
    memory limit per test256 megabytes
    inputstandard input
    outputstandard output
    Vasya has got three integers n, m and k. He’d like to find three integer points (x1,y1), (x2,y2), (x3,y3), such that 0≤x1,x2,x3≤n, 0≤y1,y2,y3≤m and the area of the triangle formed by these points is equal to nmk.

    Help Vasya! Find such points (if it’s possible). If there are multiple solutions, print any of them.

    Input
    The single line contains three integers n, m, k (1≤n,m≤109, 2≤k≤109).

    Output
    If there are no such points, print “NO”.

    Otherwise print “YES” in the first line. The next three lines should contain integers xi,yi — coordinates of the points, one point per line. If there are multiple solutions, print any of them.

    You can print each letter in any case (upper or lower).

    Examples
    inputCopy
    4 3 3
    outputCopy
    YES
    1 0
    2 3
    4 1
    inputCopy
    4 4 7
    outputCopy
    NO
    Note
    In the first example area of the triangle should be equal to nmk=4. The triangle mentioned in the output is pictured below:

    In the second example there is no triangle with area n*m/k=16/7.

    在这里插入图片描述
    不要被他给的图迷惑,这就是个水题。

    include<iostream>
    using namespace std;
    int main()
    {
       long long  int n,m,k,i;
       cin>>n>>m>>k;
      if((n*m)%(2*k)!=0)
      {
          cout<<"NO";
          return 0;
      }
      for(i=1;i<=m;i++)
      {
          if((n*i)==(2*n*m)/k)
          {
              cout<<"YES"<<endl;
              cout<<0<<' '<<0<<endl;
              cout<<0<<' '<<n<<endl;
              cout<<n<<' '<<i<<endl;
              return 0;
          }
      }
      for(i=1;i<=n;i++)
      {
          if((m*i)==(2*n*m)/k)
          {
              cout<<"YES"<<endl;
              cout<<0<<' '<<0<<endl;
              cout<<0<<' '<<m<<endl;
              cout<<0<<' '<<i<<endl;
              return 0;
          }
      }
    }
    
  • 相关阅读:
    wmware虚拟机的克隆
    解决SecureCRT无法用非root账号登录ssh
    Docker容器操作
    Docker镜像操作
    Docker的安装和启动
    linux安装tomcat
    POJ 2456 Aggressive cows ( 二分搜索)
    POJ 1064 Cable master (二分查找)
    2008 APAC local onsites C Millionaire (动态规划,离散化思想)
    贿赂囚犯 Bribe the prisoners ( 动态规划+剪枝)
  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798896.html
Copyright © 2020-2023  润新知