1042 Shuffling Machine

part 1, 1.1

自己解法

• 暴力法
• order 用于保存输入的顺序指令序列
• a 为实际进行操纵的数组
• b 为辅助操作的数组

#include <iostream>
using namespace std;
#define n 54int order[n] = {0};
int a[n] = {0};
int b[n] = {0};void func()
{for (int i = 0; i < n; i++)a[order[i] - 1] = b[i];for (int i = 0; i < n; i++)b[i] = a[i];
}int main()
{string repCards[n] = {""};string repLetter[5] = {"S", "H", "C", "D"};for (int i = 0; i < 4; i++)for (int j = 1; j <= 13; j++){int pos = i * 13 + j - 1;repCards[pos] = repLetter[i] + to_string(j);}repCards[52] = "J1";repCards[53] = "J2";int k;cin >> k;for (int i = 0; i < n; i++){cin >> order[i];a[i] = i;b[i] = i;}for (int i = 0; i < k; i++)func();cout << repCards[a[0]];for (int i = 1; i < n; i++)cout << " " << repCards[a[i]];cout << endl;system("pause");return 0;
}

大神解法

• 柳神
• 本质上和我的是一样的，scan，start，end三个数组

#include <cstdio>
using namespace std;
int main() {int cnt;scanf("%d", &cnt);int start[55], end[55], scan[55];for(int i = 1; i < 55; i++) {scanf("%d", &scan[i]);end[i] = i;}for(int i = 0; i < cnt; i++) {for(int j = 1; j < 55; j++)start[j] = end[j];for(int k = 1; k < 55; k++)end[scan[k]] = start[k];}char c[6] = {"SHCDJ"};for(int i = 1; i < 55; i++) {end[i] = end[i] - 1;printf("%c%d", c[end[i]/13], end[i]%13+1);if(i != 54) printf(" ");}return 0;
}