Q-Sequence

Q-Sequence

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 672    Accepted Submission(s): 334

Problem Description

A Q-sequence is defined as:

Q-Seq := 0 or
Q-Seq := Q-seq Q-seq 1

That is to say a Q-Sequence is a single '0' or two Q-Sequences followed by an '1'.

Given a sequence of '0's and '1's, you are to determine whether it is a Q-Sequence.

 

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of '1's and '0's. The maximum length of the sequence is 1000.

 

Output

The output contain n lines, print "Yes" if it is a Q-sequence, otherwise print "No".

 

Sample Input

3

0010011

0101

00011

 

Sample Output

Yes

No

Yes

一开始还以为要递归呢，没想出来，后来想了一下，真的水。

只要计算0和1的个数，最后判断下就差不多了，具体思路看代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
char s[1005];

void test(){
    cout<<"**"<<endl;
}
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        mem(s);
        scanf("%s",&s);
        int len=strlen(s);
        int cnt=0;
        bool prime=true;
        for(int i=0;i<len;i++){
           // test();
            if(s[0]==1){
                prime=false;
                break;
            }
            if(s[i]=='0')
                cnt++;
            else if(s[i]=='1'&&cnt>1){
                cnt--;
            }else if(s[i]=='1'&&cnt==1){
               // test();
                prime=false;
                break;
            }
        }
        if(prime&&cnt==1){
            cout<<"Yes"<<endl;
        }else
            cout<<"No"<<endl;
    }
    return 0;
}