CF451C Predict Outcome of the Game 水题

Codeforces Round #258 (Div. 2)

Predict Outcome of the Game

 

C. Predict Outcome of the Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Sample test(s)

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).

 

题意：已知ABC 3个队已经打了k场比赛，一共要打n场比赛，已知之前AB的胜场数的差的绝对值、BC的胜场数的差的绝对值，求最后是否有可能三个队胜场相同。（每单场比赛必定会决出胜负，不会平）

题解：水题，就3个队，一共就几种情况，可以枚举判断。

·CF要的就是又快又稳，这种水题就是要怒枚举一发。已知两个绝对值，那么三个队的分数分布有4种情况，按升降来说明大概是“//" "/\" "\/" "\\"四种，定好了大小，然后根据已打场次k调整一下，就能得到4种已知胜场，看看能不能填平。

看代码可以发现我分了两个函数，非常专业。（？

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usint unsigned int
#define mz(array) memset(array, 0, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) printf("%d\n",x);
#define RE  freopen("D.in","r",stdin)
#define WE  freopen("1biao.out","w",stdout)



ll n,k,d1,d2;
bool gank(const ll &x,const ll &y,const ll &z){
    ll a[3];
    a[0]=x;
    a[1]=y;
    a[2]=z;
    sort(a,a+3);
    if(a[0]+a[1]+a[2]<k){
        ll t=(k-a[0]-a[1]-a[2])/3;
        a[0]+=t,a[1]+=t,a[2]+=t;
    }
    if(a[0]<0){a[2]-=a[0];a[1]-=a[0];a[0]-=a[0];}
    if(a[0]+a[1]+a[2]!=k)return 0;
    ll need=a[2]-a[0]+a[2]-a[1];
    if(need>n-k)return 0;
    if((n-k-need)%3!=0)return 0;
    return 1;
}

bool farm(){
    if(gank(0,d1,d1+d2))return 1;
    if(gank(0,d1,d1-d2))return 1;
    if(gank(d1,0,d2))return 1;
    if(gank(d1+d2,d2,0))return 1;
    return 0;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%I64d%I64d%I64d%I64d",&n,&k,&d1,&d2);
        if(farm())puts("yes");
            else puts("no");
    }
    return 0;
}