HDU-1023-Train Problem I

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10986    Accepted Submission(s): 5820

Problem Description

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

 

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

 

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

 

Sample Input

12310

 

Sample Output

12516796

Hint

The result will be very large, so you may not process it by 32-bit integers.

 

Author

Ignatius.L

思路:都写在代码注释上了

代码:

#include<iostream>
#include<stack>
#include<cstdio>
#include<string.h>
using namespace std;
int vis[50];
char str1[50],str2[50];
int main()
{
	int n,i,j,s,vis[50];
	while(~scanf("%d %s%s",&n,str1,str2)){
		//cout<<str1<<endl<<str2<<endl;
		stack<char>st;//选择在此处写一个stack,则可以不用进行清空，亲测可以节省时间 
		memset(vis,0,sizeof(vis));
		j=s=0;
		for(i=0;i<n;i++){
			st.push(str1[i]);
			vis[s++]=1;
			while(!st.empty()&&st.top()==str2[j]){//如果栈顶元素和字符串2的首元素相等，则将栈顶元素抛出 
				vis[s++]=0;                       //同时字符串2后移一位(字符串2对应的下标只能往后移动)
				st.pop();
				j++;
			}
		}
		if(j==n){
			printf("Yes.\n");
			for(i=0;i<s;i++){
				if(vis[i]) printf("in\n");
				else printf("out\n");
			}
		}
		else{
			printf("No.\n");
		}
		printf("FINISH\n");
	}
	return 0;
}