解题思路：
给出 n m 牌的号码是从1到n*m 你手里的牌的号码是1到n*m之间的任意n个数，每张牌都只有一张，问你至少赢多少次
可以转化为你最多输max次，那么至少赢n-max次 而最多输max次，则是对方最多赢max次，则用对方的最小的牌去依次比较你手中的牌（按照升序排），如果找到有比它小的，则对方赢一次 依次循环直到遍历完对方的牌。
Game Prediction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 818 Accepted Submission(s): 453
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1010],b[1010],used[1010];
int main()
{
int n,m,i,j,sum,flag=1;
while(scanf("%d %d",&m,&n)!=EOF&&(n||m))
{
sum=0;
memset(a,1,sizeof(a));
memset(used,1,sizeof(used));
for(i=1;i<=n;i++)
{
scanf("%d",&b[i]);
a[b[i]]=0;
}
sort(b,b+n);
for(i=1;i<=n*m;i++)
{
if(a[i])
{
for(j=1;j<=n;j++)
{
if(i>b[j]&&used[j])
{
sum++;
used[j]=0;
break;
}
}
}
}
printf("Case %d: %d
",flag,n-sum);
flag++;
}
}