zoukankan      html  css  js  c++  java
  • CodeForces 451C Predict Outcome of the Game

    Predict Outcome of the Game
    Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

    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
    ",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
    ",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
    ",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
    ",4);
    74         }
    75         
    76 
    77         if(flg==1)
    78             printf("yes
    ");
    79         else
    80             printf("no
    ");
    81     }
    82 
    83 }
    View Code
  • 相关阅读:
    paip.云计算以及分布式计算的区别
    paip.索引的种类以及实现attilax 总结
    paip.分布式应用系统java c#.net php的建设方案
    paip.提升性能--多核编程中的java .net php c++最佳实践 v2.0 cah
    paip.中文 分词 ---paoding 3.1 的使用
    paip.2013年技术趋势以及热点 v2.0 cae
    paip.为什么使用多线程的原因.
    paip.提升性能--多核cpu中的java/.net/php/c++编程
    paip.重装系统需要备份的资料总结..v2.0 cad
    paip.禁用IKAnalyzer 的默认词库.仅仅使用自定义词库.
  • 原文地址:https://www.cnblogs.com/cyd308/p/4771554.html
Copyright © 2011-2022 走看看