zoukankan      html  css  js  c++  java
  • The Power of Xiangqi(水题)

    The Power of Xiangqi

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 284    Accepted Submission(s): 145

    Problem Description

    Xiangqi is one of the most popular two-player board games in China. The game represents a battle between two armies with the goal of capturing the enemy’s “general” piece.
    Now we introduce some basic rules of Xiangqi. Xiangqi is played on a 10×9 board and the pieces are placed on the intersections (points). There are two groups of pieces marked by black or red Chinese characters, belonging to the two players separately. During the game, each player in turn moves one piece from the point it occupies to another point. No two pieces can occupy the same point at the same time. A piece can be moved onto a point occupied by an enemy piece, in which case the enemy piece is "captured" and removed from the board. When the general is in danger of being captured by the enemy player on the enemy player’s next move, the enemy player is said to have "delivered a check". If the general's player can make no move to prevent the general's capture by next enemy move, the situation is called “checkmate”.
    Each player has 7 kinds of pieces in this game. Their names, offense power and symbol letters are shown in the table below:



    Now show you the pieces of red player and black player, you are going to find out which player has the larger total offense power. Since Ma and Pao working together can have good effect, if a player has no Ma or no Pao, or has neither, his total offense power will be decreased by one. But the total offense power can't be decreased to zero, it is at least one.
     
    Input
    The first line is an integer T ( T <= 20) meaning there are T test cases.
    For each test case: The first line shows which pieces the red player has. It begins with an integer n ( 0 < n <= 16) meaning the number of pieces.
    Then n letters follows, all separated by one or more blanks. Each letter is a symbol letter standing for a piece, as shown in the table above.
    The second line describes the situation of the black player, in the same format as the first line.
     
    Output
    For each test case, if the red player has more offense power, then print "red". If the black player has more offense power, then print "black". If there is a tie, print "tie".
     
    Sample Input
    3 2 A B 2 A B 7 A A B C D D F 7 A A B B C C F 5 A A B B F 3 A B F
     
    Sample Output
    tie black red
     
    Source
     

    Statistic | Submit | Discuss | Note


    因为大小写出错,还WA了一次。。。


    
    
     1 #include <iostream>
     2 #include <fstream>
     3 #include <string>
     4 #include <cstdio>
     5 #include <cmath>
     6 #include <cstring>
     7 #include <algorithm>
     8 #include <map>
     9 #include <vector>
    10 #include <queue>
    11 #include <stack>
    12 #define LL long long
    13 #define MAXI 2147483647
    14 #define MAXL 9223372036854775807
    15 #define eps (1e-8)
    16 #define dg(i) cout << "*" << i << endl;
    17 
    18 using namespace std;
    19 
    20 int main()
    21 {
    22     const int power[7] = {16, 7, 8, 1, 1, 2, 3};
    23     int T, n, R, B;
    24     char c;
    25     bool ma, pao;
    26     scanf("%d", &T);
    27     while(T--)
    28     {
    29         R = B = 0;
    30         scanf("%d", &n);
    31         ma = pao = 0;
    32         while(n--)
    33         {
    34             cin >> c;
    35             R += power[c - 'A'];
    36             if(c == 'C') pao = 1;
    37             else if(c == 'B') ma = 1;
    38         }
    39         if(pao + ma != 2 && R > 1) R--;
    40         scanf("%d", &n);
    41         ma = pao = 0;
    42         while(n--)
    43         {
    44             cin >> c;
    45             B += power[c - 'A'];
    46             if(c == 'C') pao = 1;
    47             else if(c == 'B') ma = 1;
    48         }
    49         if(pao + ma != 2 && B > 1) B--;
    50         if(B > R) puts("black");
    51         else if(B == R) puts("tie");
    52         else puts("red");
    53     }
    54     return 0;
    55 }


  • 相关阅读:
    leetcode(85)最大矩形
    红黑树
    查询学生成绩表中大于60分的每一个成绩的人数
    聚合函数以及SQL中的一些小知识
    Oracle的JDBC
    StringBuffer的reverse方法
    查询时报第一页没有数据,第二页有数据的异常
    普通人如何从平庸到优秀,在到卓越
    HDMI、DVI、VGA等这些接口
    显卡上的VGA接口和高清接口有什么区别?
  • 原文地址:https://www.cnblogs.com/cszlg/p/2910434.html
Copyright © 2011-2022 走看看