hdu 1338 Game Prediction

Game Prediction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1147    Accepted Submission(s): 644

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
题目没看懂，放了一会儿，然后同学给我说了题意，很简单，结果因为马虎错了两次。
我竟然给排排序。。。。。。。。 。。#include<stdio.h>#include<iostream>#include<algorithm>#include<cstring>using namespace std;int main(){   int n,m,s,T=1;   int a[1005];   while(scanf("%d%d",&n,&m)&&n&&m){        memset(a,0,sizeof(a));        for(int i=1;i<=m;i++){            scanf("%d",&s);            a[s]=1;        }        //sort(a+1,a+n*m+1);        int ans=0,count=0;        for(int i=m*n;i>0;i--){            if(!a[i]) {                 ans++;            }            else {                if(ans>0) ans--;                else count++;            }        }        printf("Case %d: %d\n",T++,count);   }}