SOJ 1018. A Card Trick

题目大意：一张卡牌由value和suit决定。value的范围(从小到大)是A,2,3,4,5,6,7,8,9,10,J,Q,K。suit的范围(从小到大)是C,D,H,S。

　　　　   两张牌比较大小：value较小的牌较小，若value值相等，suit较小的牌较小。

　　　　　给出五张牌，要求对五张牌进行排列，以满足某种要求：第一张牌和第二张牌的suit要求相同，从后三张牌中选择最小的牌，获得它的位置pos_min(1,2或3)，后三张牌除去最小的牌剩余的两张牌如果不是从小到大有序的则addition = 3,否则addition = 0. 第二张牌的value + pos_min + addition（若大于13则三者和减13）等于第一张牌的value。

解题思路：模拟。结构化牌，包括value和suit两个元素。对五张牌进行全排列，对每一个排列检查是否满足要求，找到第一个满足要求的排列即可。

代码如下：

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

class Card {
public:
    friend istream & operator>>(istream &is, Card &card) {
        string text;
        is >> text;
        // for (int i = 0; i < text.size(); ++i) text[i] = toupper(text[i]);
        card.value = atoi(text.c_str());
        // 获取value
        if (card.value == 0) {    // 没有数字开头
            if (text[0] == 'J') {
                card.value = 11;
            } else if (text[0] == 'Q') {
                card.value = 12;
            } else if (text[0] == 'K') {
                card.value = 13;
            } else if (text[0] == 'A') {
                card.value = 1;
            } else {
                while (true);
            }
        }
        // 获取suit
        /*if (text.back() == 'C') {
            card.suit = 1;
        } else if (text.back() == 'D') {
            card.suit = 2;
        } else if (text.back() == 'H') {
            card.suit = 3;
        } else if (text.back() == 'S') {
            card.suit = 4;
        }*/
        if (text[text.size()-1] == 'C') {
            card.suit = 1;
        } else if (text[text.size()-1] == 'D') {
            card.suit = 2;
        } else if (text[text.size()-1] == 'H') {
            card.suit = 3;
        } else if (text[text.size() - 1] == 'S') {
            card.suit = 4;
        } else {
            while (true);
        }
        return is;
    }

    friend ostream & operator<<(ostream &os, const Card &card) {
        if (card.value == 1) os << "A";
        else if (card.value < 11) os << card.value;
        else if (card.value == 11) os << "J";
        else if (card.value == 12) os << "Q";
        else if (card.value == 13) os << "K";
        if (card.suit == 1) os << "C";
        else if (card.suit == 2) os << "D";
        else if (card.suit == 3) os << "H";
        else if (card.suit == 4) os << "S";
        return os;
    }

    bool operator<(const Card &other) const {
        return (value < other.value) || (value == other.value && suit < other.suit);
    }
    int value;
    int suit;
};

int main() {
    int times;
    const int maxn = 5;
    Card cards[maxn];
    scanf("%d", &times);
    for (int t = 1; t <= times; ++t) {
        for (int i = 0; i < maxn; ++i) cin >> cards[i];
        sort(cards, cards + maxn);
        do {
            if (cards[0].suit != cards[1].suit) continue;
            int addition;
            if (cards[2] < cards[3] && cards[2] < cards[4]) {    // cards[2] 最小
                addition = 1;
                addition += cards[3] < cards[4] ? 0 : 3;
            } else if (cards[3] < cards[2] && cards[3] < cards[4]) {    // cards[3] 最小
                addition = 2;
                addition += cards[2] < cards[4] ? 0 : 3;
            } else if (cards[4] < cards[2] && cards[4] < cards[2]) {
                addition = 3;
                addition += cards[2] < cards[3] ? 0 : 3;
            }
            if (cards[0].value == (cards[1].value + addition - 1) % 13 + 1) break;
        } while (next_permutation(cards, cards + maxn));
        cout << "Problem " << t << ": ";
        for (int i = 0; i < maxn; ++i) {
            cout << cards[i] << (i + 1 == maxn ? "
" : " ");
        }
    }
    return 0;
}