Game Prediction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1584    Accepted Submission(s): 891

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 51 7 2 10 96 1162 63 54 66 65 61 57 56 50 53 480 0

 

Sample Output

Case 1: 2Case 2: 4

 

Source

Asia 2002, Beijing (Mainland China)

题意：m个人，每个人n张牌。一共有m*n张牌（点数从1,2,3.....n）。现给出你的n张牌（数组a）。求其最少能赢的次数。

思路：一共有m*n个牌，无重复。在m*n中，把除了你出现的n张牌标记出来，转存到数组b。即求b数组能大于a数组的个数。首先对a，b数组从大到小排序。设置tt作为b数组的下标，初始化为0.for循环对a数组遍历。如果有b[k]>a[i],sum++;结果为n-sum。

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;int n,m;int vis[1002];int cmp(int a,int b){return a>b;}int main(){int Case=1;while(scanf("%d%d",&m,&n)!=EOF){if(m==0&&n==0)break;int a[55];int b[55];memset(vis,0,sizeof(vis));memset(a,0,sizeof(a));memset(b,0,sizeof(b));for(int i=0;i<n;i++){scanf("%d",&a[i]);vis[a[i]]=1;}//获取b个小于a[i]的最大数int k=0;for(int i=m*n;i>=0;i--){if(k==n)break;if(!vis[i])b[k++]=i;}/*for(int i=0;i<n;i++)printf("%d  ",b[i]);printf("\n");*/sort(a,a+n,cmp);//由大到小 sort(b,b+n,cmp);/*for(int i=0;i<n;i++)printf("%d  ",a[i]);printf("\n");*/int tt=0;int sum=0;for(int i=0;i<n;i++){if(a[i]>b[tt])continue;else {tt++;sum++;}}printf("Case %d: %d\n",Case++,n-sum);}}