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

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解題報告

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

題意：

n場比賽其中k場是沒看過的，對於這k場比賽，a,b,c三隊贏的場次的關係是a隊與b隊的絕對值差d1,b隊和c隊絕對值差d2,求是否能使三支球隊的贏的場次相同。

思路：

|B-A|=d1

|C-B|=d2

A+B+C=k

這樣就有4種情況，分別是：

B>A&&C<B

B>A&&C>B

B<A&&C<B

B<A&&C>B

分別算出在k場比賽中a,b,c三支隊伍贏的場次，另外n-k場比賽分別給3支隊伍加上，看看是否能相同。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define LL long longusing 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 Gametime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard 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

53 0 0 03 3 0 06 4 1 06 3 3 03 3 3 2

output

yesyesyesnono

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).