HDU 1338 Game Prediction 贪心

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)

 

题目意思：

有m个人，每个人有n张牌，牌点为在1~n*m中的不同的数。每回合每个人出一张牌，点数最大的那个人赢，给出A人初始时的n张牌的牌点，问A至少赢的次数。

思路： 看做两个人单挑，你有m张牌，我有m*(n-1)张牌，每次我都出比你大大一点的牌，如果没有，出最小的m张牌（可以忽略），不用担心什么的

上代码了：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 10005int n,m;int vis[N],a[N];int main(){    int i,ca=0;    while(~scanf("%d%d",&n,&m),n+m)    {        memset(vis,0,sizeof(vis));        for(i=0;i<m;i++)        {            scanf("%d",&a[i]);            vis[a[i]]=1;      //我手上没有的牌        }        int ans=0;        sort(a,a+m);  //排序对方的牌        int top;                 for(i=m-1;i>=0;i--)  //这里for(i=0;i<m;i++) 也是一样正确的，不必纠结        {            for(top=a[i];;top++)                if(!vis[top])                 break;            if(top<=n*m)                vis[top]=1;  //出掉这张牌            else                ans++;    //没有比他大的，出最小牌（可以忽略出最小牌）        }        printf("Case %d: %d\n",++ca,ans);    }    return 0;}