问题描述:略
解决方法:按照魔术进行的顺序模拟即可
AC代码
#include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<string> #pragma warning(disable:4996) #define max(x,y) (x)>(y)?(y):(y) #define min(x,y) (x)<(y)?(y):(y) #define INF Ox3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 typedef long long ll; using namespace std; int main() { int n, order[100]; string card[100], word[100]; //card存储牌名,word存储命令 while (cin >> n && n) //order模拟队列 { for (int i = 0; i < n; i++) { cin >> card[i] >> word[i]; order[i] = -1; //order[i]=-1表示i位置没有牌 } for (int i = 0, j = 0; i < n; i++) { //i表示现在要放第i个牌,j记录位置 for (int k = word[i].length();; j = (j + 1) % n) //利用模除实现循环 if (order[j] == -1 && --k == 0) break; order[j] = i; //j位置上放第i张牌 } for (int i = 0; i < n; i++) //输出 if(i) cout<<' '<< card[order[i]]; else cout << card[order[i]]; cout << endl; } }