CF451C Predict Outcome of the Game 水题-优快云博客

本文介绍了一个简单的算法，用于预测在已知部分比赛结果的情况下，三支足球队伍在一系列比赛中是否可能最终胜场相同，确保没有队伍赢得整个锦标赛。

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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

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。（？

 1 //#pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<map>
 9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll long long
14 #define usint unsigned int
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(i=0;i<(n);i++)
18 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) printf("%d\n",x);
23 #define RE  freopen("D.in","r",stdin)
24 #define WE  freopen("1biao.out","w",stdout)
25 
26 
27 
28 ll n,k,d1,d2;
29 bool gank(const ll &x,const ll &y,const ll &z){
30     ll a[3];
31     a[0]=x;
32     a[1]=y;
33     a[2]=z;
34     sort(a,a+3);
35     if(a[0]+a[1]+a[2]<k){
36         ll t=(k-a[0]-a[1]-a[2])/3;
37         a[0]+=t,a[1]+=t,a[2]+=t;
38     }
39     if(a[0]<0){a[2]-=a[0];a[1]-=a[0];a[0]-=a[0];}
40     if(a[0]+a[1]+a[2]!=k)return 0;
41     ll need=a[2]-a[0]+a[2]-a[1];
42     if(need>n-k)return 0;
43     if((n-k-need)%3!=0)return 0;
44     return 1;
45 }
46 
47 bool farm(){
48     if(gank(0,d1,d1+d2))return 1;
49     if(gank(0,d1,d1-d2))return 1;
50     if(gank(d1,0,d2))return 1;
51     if(gank(d1+d2,d2,0))return 1;
52     return 0;
53 }
54 
55 int main(){
56     int T;
57     scanf("%d",&T);
58     while(T--){
59         scanf("%I64d%I64d%I64d%I64d",&n,&k,&d1,&d2);
60         if(farm())puts("yes");
61             else puts("no");
62     }
63     return 0;
64 }