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