Predict Outcome of the Game

CodeForces - 451C

There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.

You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.

You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?

Note that outcome of a match can not be a draw, it has to be either win or loss.

Input

The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).

Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.

Output

For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).

Examples

Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no

Note

Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.

Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".

Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).

sol:大力分类讨论四种情况,要及其仔细,否则会挂的很惨。。。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
ll s=;
bool f=;
char ch=' ';
while(!isdigit(ch))
{
f|=(ch=='-'); ch=getchar();
}
while(isdigit(ch))
{
s=(s<<)+(s<<)+(ch^); ch=getchar();
}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<)
{
putchar('-'); x=-x;
}
if(x<)
{
putchar(x+''); return;
}
write(x/);
putchar((x%)+'');
return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
ll T,n,m,d1,d2;
#define PYES {puts("yes"); continue;}
#define PNO {puts("no"); continue;}
int main()
{
R(T);
while(T--)
{
R(n); R(m); R(d1); R(d2);
if(n%) PNO;
ll tmp=n/,oo;
if(max(d1,d2)>tmp) PNO
if(d1>=d2) //0,d1,d1-d2
{
oo=m-d1-d1+d2;
if((d1+oo/)<=tmp&&oo>=&&oo%==) PYES
}
else //d2-d1,d2,0
{
oo=m-d2-d2+d1;
if((d2+oo/)<=tmp&&oo>=&&oo%==) PYES
}
oo=m-d1-d2; //d1,0,d2
if((max(d1,d2)+oo/)<=tmp&&oo>=&&oo%==) PYES
oo=m-d1-d2-d2; //d1+d2,d2,0
if((d1+d2+oo/)<=tmp&&oo>=&&oo%==) PYES
oo=m-d1-d1-d2; //0,d1,d1+d2
if((d1+d2+oo/)<=tmp&&oo>=&&oo%==) PYES
PNO
}
return ;
}
/*
Input
8
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
3 1 1 1
9 4 1 2
3 3 1 1
Output
yes
yes
yes
no
no
yes
yes
no
*/

codeforces451C的更多相关文章

随机推荐

  1. input中只能写入数字int、float

    input 属性 type="number" <input type="number" min="1" max="100&q ...

  2. 认证与Shiro安全框架

    本文内容均来自官网 1.简介 Apache Shiro是Java的一个安全框架.功能强大,使用简单的Java安全框架,它为开发人员提供一个直观而全面的认证,授权,加密及会话管理的解决方案. 实际上,S ...

  3. 【Dojo 1.x】笔记目录

    学习笔记和教程是不同的,笔记是随心记,学到什么就写什么,我尽量按逻辑顺序写笔记. Dojo是什么? Dojo是这么一个JavaScript框架,区别于jQuery等小型类库,这个类库更合适于构建Web ...

  4. Android为TV端助力 UDP协议

    废话不多说.直接上代码! 一. 接收端 1.创建UDP连接 public void init() { try { //开关的作用 isRunning = true; DatagramSocket mU ...

  5. 微信小程序 从含有tabbar的页面跳转到不含有tabbar的页面

    如何离开含有tabbar的页面 在微信小程序开发过程中,我们会碰到从某页跳转到一个含有tabbar的页面的需求, 用 wx.navigateTo({url: '...',})  不起作用,需要使用 w ...

  6. 从分治算法到 Hadoop MapReduce

    从分治算法说起 要说 Hadoop MapReduce 就不得不说分治算法,而分治算法其实说白了,就是四个字 分而治之 .其实就是将一个复杂的问题分解成多组相同或类似的子问题,对这些子问题再分,然后再 ...

  7. 手机Soc芯片简介

    手机SoC(System On a Chip,在一个芯片里面集成CPU.GPU.SP.ISP.RAM内存.Wi-Fi控制器.基带芯片以及音频芯片等)芯片(基于arm架构指令集) 高通骁龙(Snapdr ...

  8. Python函数二(函数名,闭包,迭代器)之杵臼之交

    函数名的使用: 函数名可以作为值,赋值给变量. 函数名可以作为参数传参给函数. 函数名可以作为返回值. 函数名可以作为元素存储在容器里. 闭包:在嵌套函数内,使用外层局部变量(非全局变量)就是一个闭包 ...

  9. Linux shell 及命令汇总

    1 文件管理命令 1.cat命令:将文件内容连接后传送到标准输出或重定向到文件 2.chmod命令:更改文件的访问权限 3.chown命令:更改文件的所有者 4.find命令:查找(符合条件)文件并将 ...

  10. Python开发【内置模块篇】datetime

    获取当前日期和时间 >>> from datetime import datetime >>> now = datetime.now() >>> ...