PAT 1042 Shuffling Machine

#include <cstdio>
#include <cstdlib>
#include <vector>

using namespace std;

char tbl[5] = {‘S‘, ‘H‘, ‘C‘, ‘D‘, ‘J‘};

void shuffle(vector<char> &card, vector<char> &rnd) {
    int rlen= rnd.size();
    vector<char> tmp(card.size(), 0);
    for (int i=0; i<rlen; i++) {
        tmp[rnd[i]] = card[i];
    }
    card = tmp;
}

void print_card(char card) {
    printf("%c%d", tbl[card/13], card % 13 + 1);
} 

int main() {
    int times = 0, r = 0;
    vector<char> rnd(54, 0);
    vector<char> card(54, 0);
    for (int i=0; i<54; i++) {
        card[i] = i;
    }
    scanf("%d", &times);
    for (int i=0; i<54; i++) {
        scanf("%d", &r);
        rnd[i] = r - 1;
    }
    for (int i=0; i<times; i++) {
        shuffle(card, rnd);
    }
    print_card(card[0]);
    for (int i=1; i<54; i++) {
        printf(" ");
        print_card(card[i]);
    }
    return 0;
}

一开始想复杂了以为要像算法导论里面提到的那样进行元素交换,那里是因为随机数列不能保证在一定范围内唯一(如果要得到这样的数列其实也可以,就相当于已经进行了一次洗牌)所以遍历数组时产生的随机数是多少就把当前元素和下标和当前随机数一致的元素对换。不过这里已经说了数列是不重复的就直接把元素放到对应的位置上即可。当然程序里可以不把单次shuffle写成一个函数,这样可以避免反复创建vector,只要使用一个临时vector即可,多执行几次swap来代替元素拷贝。

时间: 2024-12-24 19:45:42

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博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789205.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~ 给出洗牌次数,以及洗牌的序列规则第i个数shuffles[i]表示要将第i张牌移到第shuffles[i]个 很简单,就是shuffle_seq[shuffles[i]]=start_seq[i]; #include <iostream> #include