1054Game Prediction小结

1.题目描述：

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

2.题目概述及思路：M个人，每个人N张牌，牌的点数从M*N向下，没有重复的点数。每次出一张牌，谁的点大，谁这轮就赢了。给你一套牌，看最少能赢多少轮。思路很简单，去除掉给出的点数，然后构造一套最好的牌。

3.具体实现：

#include <iostream>#include <string>using namespace std;//练习下快忘了的快排void quick_sort(int s[], int l, int r){    if (l < r)    {        int i = l, j = r, x = s[l];        while (i < j)        {            while(i < j && s[j] <= x)                j--;            if(i < j)                s[i++] = s[j];            while(i < j && s[i] > x)                i++;            if(i < j)                s[j--] = s[i];        }        s[i] = x;        quick_sort(s, l, i - 1);        quick_sort(s, i + 1, r);    }}int main(){    int r1[60],r2[60];    int m,n,max,i,j;    int count=0;    int cas=1;    while(true)    {        cin>>m>>n;        if(m==0&&n==0) break;        for(i=0; i<n; i++)            cin>>r1[i];        quick_sort(r1,0,n-1);        for(i=0; i<60; i++)            r2[i]=0;        int flag=0;        i=0;        count=0;        max=m*n;        while(i<n)        {            flag=0;            for(j=0; j<n; j++)            {                if(r1[j]==max) flag=1;            }            if(flag==0)            {                r2[i]=max;                i++;            }            max--;        }        i=0;        count=0;        j=0;        for(i=0; i<n; i++)        {            if(r1[i]>r2[j])            {                count++;            }            else            {                j++;            }        }        cout<<"Case "<<cas<<": "<<count<<endl;        cas++;    }    return 0;}