1042 Shuffling Machine （20 分）

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最新推荐文章于 2022-03-24 15:13:54 发布

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本文详细解析了1042ShufflingMachine题目中的模拟洗牌算法，通过定义新的字符串数组作为中间产物来临时存放卡牌，实现每次洗牌后的卡牌位置更新。代码示例展示了如何使用C++实现这一过程。

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题目地址：1042 Shuffling Machine （20 分）

模拟题，定义一个新的字符串数组作为中间产物临时存放卡牌。每执行一次，中间数组更新一次，然后在复制到卡牌数组中。

代码

#include <bits/stdc++.h>
using namespace std;
string a[] = {"0","S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13",
			  "H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13",
			  "C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13",
			  "D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13",
			  "J1","J2"
};
string s[55];
int main() {
	//freopen("datain.txt","r",stdin);
	int n;
	scanf("%d", &n);

	int next[55];//临时数组
	next[0] = 0;
	for (int i = 1; i < 55; i++ ) {
		scanf("%d", &next[i]);
	}
	while (n--) {
		for (int i = 1; i < 55; i++) {
			s[next[i]] = a[i];
		}
		
		for (int i = 1; i < 55; i++) {
			a[i] = s[i];
		}
	}
	cout<<a[1];
	for (int i = 2; i < 55; i++) {
		cout<<" "<<a[i];
	}
	cout<<endl;
	return 0;
}