Game Prediction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Suppose there are M people, including you, playing a special card game. At the beginning, each player receives N cards. The pip of a card is a positive integer which is at most N*M. And there are no two cards with the same pip. During a round, each player chooses
one card to compare with others. The player whose card with the biggest pip wins the round, and then the next round begins. After N rounds, when all the cards of each player have been chosen, the player who has won the most rounds is the winner of the game.

Given your cards received at the beginning, write a program to tell the maximal number of rounds that you may at least win during the whole game.

Input

The 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 players and the number of cards each player 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.

Output

For 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

Source

Asia 2002, Beijing (Mainland China)

通过已给出的数组 a  找出其余所有人的卡牌存放在一个数组 b 中

通过求 b 能赢得最多场max来间接求出 a 能赢得最少场ans

ans=n-max

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int a[55],b[1005];
    int i,j,ans,k,l;
    int n,m,t=0;
    while(cin>>n>>m,m&&n)
    {
        t++;
        memset(b,0,sizeof(b));
        for(i=0;i<m;i++)
            cin>>a[i];
        sort(a,a+m,cmp);
        /*for(i=0;i<m;i++)
        cout<<a[i];*/
        l=0;
        ans=0;
        for(i=m*n;i>a[m-1];i--)
        {   k=0;
            for(j=0;j<m;j++)
             {
                if(a[j]==i)
                   {
                        k=1;
                    continue;
                   }
             }
             if(!k)
             {b[l]=i;/*cout<<b[l]<<" ";*/l++;}
        }
        k=m*(n-1);
        j=0;
        for(i=0;i<m;i++)
           {
              if(a[i]<b[j])
              {
                  ans++;
                  j++;
              }
           }
           cout<<"Case "<<t<<": ";
        cout<<(m-ans)<<endl;
    }

    return 0;
}

hdu 1338Game Prediction