Codeforces Round #258 (Div. 2/C)/Codeforces451C_Predict Outcome of the Game(枚举)

解题报告

http://blog.csdn.net/juncoder/article/details/38102391

题意：

n场比赛其中k场是没看过的，对于这k场比赛，a,b,c三队赢的场次的关系是a队与b队的绝对值差d1,b队和c队绝对值差d2,求是否能使三支球队的赢的场次相同。

思路：

|B-A|=d1

|C-B|=d2

A+B+C=k

这样就有4种情况，分别是：

B>A&&C

B>A&&C>B

B

BB

分别算出在k场比赛中a,b,c三支队伍赢的场次，另外n-k场比赛分别给3支队伍加上，看看是否能相同。

#include 
#include 
#include 
#include 
#define LL long long
using namespace std;

int main()
{
    int t,i,j;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            LL d1,d2,n,k,a,b,c;
            scanf("%lld%lld%lld%lld",&n,&k,&d1,&d2);
            int f=0;
            LL kk=n/3;
            //1
            double fa=(double)((k+d2)-2*d1)/3;
            if(fa>=0&&fa==(LL )fa)
            {
                a=(LL)fa;
                b=d1+a;
                c=b-d2;
                if(a>=0&&b>=0&&c>=0&&b<=kk&&c<=kk&&a<=kk&&(kk-b+kk-a+kk-c)==(n-k))
                {
                    f=1;
                }
            }
            //2
            fa=(double)((k-d2)-2*d1)/3;
            if(fa>=0&&fa==(LL )fa)
            {
                a=(LL)fa;
                b=d1+a;
                c=b+d2;
                if(a>=0&&b>=0&&c>=0&&b<=kk&&c<=kk&&a<=kk&&(kk-b+kk-a+kk-c)==(n-k))
                {
                    f=1;
                }
            }
            //3
            fa=(double)((k+d2)+2*d1)/3;
            if(fa>=0&&fa==(LL )fa)
            {
                a=(LL )fa;
                b=a-d1;
                c=b-d2;
                if(a>=0&&b>=0&&c>=0&&b<=kk&&c<=kk&&a<=kk&&(kk-b+kk-a+kk-c)==(n-k))
                {
                    f=1;
                }
            }
            //4
            fa=(double)((k-d2)+2*d1)/3;
            if(fa>=0&&fa==(LL )fa)
            {
                a=(LL)fa;
                b=a-d1;
                c=b+d2;
                if(a>=0&&b>=0&&c>=0&&b<=kk&&c<=kk&&a<=kk&&(kk-b+kk-a+kk-c)==(n-k))
                {
                    f=1;
                }
            }
            if(f==1)
                printf("yes\n");
            else printf("no\n");
        }
    }
    return 0;
}

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 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 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).