Game Prediction

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最新推荐文章于 2022-12-02 22:29:21 发布

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分类专栏：我的ACM 文章标签： each input integer output

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本文探讨了一种特殊卡牌游戏的策略算法，目标是在游戏中获胜最多轮次。该算法考虑了玩家数量、每轮发出的卡牌数及卡牌数值等因素。通过排序和策略选择，确保玩家能在有限的轮次中取得最优成绩。

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Time limit: 1 Seconds Memory limit: 32768K

Total Submit: 149 Accepted Submit: 112

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 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

题目很简单，但是代码没有通过，纠结中……

#include <iostream> using namespace std; int card[60]; int used[1050]; int main() { int m, n, i, j; int win, flag = 0; while (cin>>m>>n) { if (m == 0 && n == 0) break; for (i = 1; i <= m*n; ++i) used[i] = 0; for (i = 1; i <= n; ++i) { cin>>card[i]; used[card[i]] = 1; } for (i = 1; i < n; ++i) { for (j = i+1; j <= n; ++j) { if (card[i] < card[j]) { int tmp = card[i]; card[i] = card[j]; card[j] = tmp; } } } win = 0; i = 1; for (j = m*n; j > 0 && i <= n; --j) { if (used[j] == 1) { ++win; ++i; } else if (used[j] == 0) { used[card[i]] = 2; ++i; } } cout<<"Case "<<++flag<<":"<<win<<endl; } return 0; }