POJ 1323 Game Prediction

Game Prediction

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10627		Accepted: 5121

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?20) and n (1?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

Beijing 2002

 

 

 

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int card[25];
int main()
{
    int m,n,a,ca=1,cm,ans;
    while(scanf(" %d%d",&m,&n)==2&&n&&m)
    {
        for(int i=0;i<n;i++){
            scanf("%d",&card[i]);
        }
        sort(card,card+n);
        cm=n*m; ans=0;
        for(int i=n-1;i>=0;i--)
        {
            if(cm==card[i]) ans++,cm--; //当前最大的卡在自己手里，必胜
            else cm-=2;//对方胜，下一轮要获胜的话，卡的值要是当前最大值-2，因为没有重复的卡。
        }
        printf("Case %d: %d\n",ca++,ans);
    }
}