Game Prediction（贪心）

最新推荐文章于 2022-12-02 22:29:21 发布

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本文介绍了一种特殊卡牌游戏的预测算法，玩家需在每轮选择一张牌与其他玩家比较大小，最终获胜轮次最多者为胜者。算法通过排序和遍历策略计算玩家至少能赢得的最大轮数。

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Game Prediction
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Status
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

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<queue>
using namespace std;
const int maxn=55;
bool cmp(int x,int y)
{
    return x>y;
}
int main()
{
    int M;//人数
    int N;//牌数
    int casen=1;
    while(~scanf("%d%d",&M,&N)&&M&&N)
    {
        int Card[maxn];//存储自己的牌
        for(int i=0; i<N; i++)
            scanf("%d",&Card[i]);
        sort(Card,Card+N,cmp);
        int point=0;
        int result=0;
        for(int i=0,j=M*N; i<N&&j>0; j--) //遍历M*N到1
        {
            if(j==Card[i])
            {
                i++;
                point==0?result++:point--;
            }
            else
                point++;
        }
        printf("Case %d: %d\n",casen++,result);
    }
    return 0;
}