hdu-1338 game predictions（贪心题）

InputThe input consists of several test cases. The first line of each case contains two integers m (2 <= m <= 20) and n (1 <= n <= 50), representing the number of p s and the number of cards each p receives at the beginning of the game, respectively. This followed by a line with n positive integers, representing the pips of cards you received at the beginning. Then a blank line follows to separate the cases. 

The input is terminated by a line with two zeros. 
OutputFor each test case, output a line consisting of the test case number followed by the number of rounds you will at least win during the game. 
Sample Input

2 5
1 7 2 10 9

6 11
62 63 54 66 65 61 57 56 50 53 48

0 0

Sample Output

Case 1: 2
Case 2: 4

题意：模拟打牌，m个人，每人n张牌。点数1-n*m，出得最大的那个人赢一局，问你最多赢几局

题解：贪心，我出大的，对面出最小的，我出小的，对面最小但比我大的

#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
#define PI 3.14159265358979323846264338327950
bool a[1050];
int main()
{
    int m=0,n=0,count=0,t;
    while(scanf("%d %d",&m,&n) && (m||n))
    {
        count++;
        int i=0;
        int win=0,big=0;
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
        {
            scanf("%d",&t);
            a[t]=1;
        }
        for(i=n*m;i>0;i--)
        {
            if (a[i])
            {
                if (big==0)
                    ++win;
                else
                    big--;
            }
            else
                big++;
        }
        printf("Case %d: %dn",count,win);
    }
    return 0;
}