CodeForces 451C Predict Outcome of the Game

Predict Outcome of the Game

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 kgames. 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 Input

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

Hint

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).

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     int T;
 9     long long n,k,d1,d2;
10     long long i,j;
11     scanf("%d",&T);
12     while(T--)
13     {
14         scanf("%I64d %I64d %I64d %I64d",&n,&k,&d1,&d2);
15         long long x,y,z,c,v,ma;
16 
17         int flg=0;
18 
19         v=k-d1-2*d2;
20         if(v>=0 && v%3==0 && flg==0)
21         {
22             x=v/3+d1+d2,y=v/3+d2,z=v/3;
23             if((n-k-(x-y+x-z))>=0 && (n-k-(x-y+x-z))%3==0)
24                 flg=1;
25             //if(flg==1)
26             //printf("%d\n",1);
27         }
28 
29         v=k-d1+2*d2;
30         if(v>=0 && v%3==0 && flg==0)
31         {
32             x=v/3+d1-d2,y=v/3-d2,z=v/3;
33             if(x>=0 && y>=0 && z>=0)
34             {
35                 ma=max(x,y);
36                 ma=max(ma,z);
37                 if((n-k-(ma-x+ma-y+ma-z))>=0 && (n-k-(ma-x+ma-y+ma-z))%3==0)
38                     flg=1;
39             }
40             //if(flg==1)
41             //printf("%d\n",2);
42         }
43         
44 
45         v=k+d1-2*d2;
46         if(v>=0 && v%3==0 && flg==0)
47         {
48             x=v/3-d1+d2,y=v/3+d2,z=v/3;
49             if(x>=0 && y>=0 && z>=0)
50             {
51                 ma=max(x,y);
52                 ma=max(ma,z);
53                 if((n-k-(ma-x+ma-y+ma-z))>=0 && (n-k-(ma-x+ma-y+ma-z))%3==0)
54                     flg=1;
55             }
56             //if(flg==1)
57             //printf("%d\n",3);
58         }
59         
60 
61         v=k+d1+2*d2;
62         if(v>=0 && v%3==0 && flg==0)
63         {
64             x=v/3-d1-d2,y=v/3-d2,z=v/3;
65             if(x>=0 && y>=0 && z>=0)
66             {
67                 ma=max(x,y);
68                 ma=max(ma,z);
69                 if((n-k-(ma-x+ma-y+ma-z))>=0 && (n-k-(ma-x+ma-y+ma-z))%3==0)
70                     flg=1;
71             }
72             //if(flg==1)
73             //printf("%d\n",4);
74         }
75         
76 
77         if(flg==1)
78             printf("yes\n");
79         else
80             printf("no\n");
81     }
82 
83 }

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转载于:https://www.cnblogs.com/cyd308/p/4771554.html