Codeforces Round #484 (Div. 2) A. Row

Posted 2020-11-07 moxin0509

A. Row

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You‘re given a row with $\mathrm{nn chairs.\; We\; call\; a\; seating\; of\; people\; "maximal"\; if\; the\; two\; following\; conditions\; hold:}$

1. There are no neighbors adjacent to anyone seated.
2. It‘s impossible to seat one more person without violating the first rule.

The seating is given as a string consisting of zeros and ones ($\mathrm{00 means\; that\; the\; corresponding\; seat\; is\; empty, 11 —\; occupied).\; The\; goal\; is\; to\; determine\; whether\; this\; seating\; is\; "maximal".}$

Note that the first and last seats are not adjacent (if $\mathrm{n\ne 2n\ne 2).}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 10001\le n\le 1000) —\; the\; number\; of\; chairs.}$

The next line contains a string of $\mathrm{nn characters,\; each\; of\; them\; is\; either\; zero\; or\; one,\; describing\; the\; seating.}$

Output

Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".

You are allowed to print letters in whatever case you‘d like (uppercase or lowercase).

Examples

input

Copy

3
101

output

Copy

Yes

input

Copy

4
1011

output

Copy

No

input

Copy

5
10001

output

Copy

No

Note

In sample case one the given seating is maximal.

In sample case two the person at chair three has a neighbour to the right.

In sample case three it is possible to seat yet another person into chair three.

【题意】：

题意就是1代表这个位置有坐人，0代表这个位置没有坐人。

如果1旁边有坐人，就是No

如果空位置还能放下一个1，也是No

【思路】：

我发现如果有3个连续的0的话，肯定是No

如果有两个连接的1的话，肯定也是No

所以我一开始写了一个判断3个0的，然后wa

后来想了一下，两个连续的0也可能导致No

比如，001 or 100

然后我发现了一下就是一开始两个0或者是结尾两个0，

这种情况就没其他的。

特判一下就可以了；

但是还是wa。

然后我想会不会是一个数的时候出现问题

果然一个0也有可能是No

针对于0

这种情况

然后考虑完后一发ac

附上代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN = 1005;
char team[MAXN];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {//printf("...\n");
        memset(team,0,sizeof(team));
        scanf("%s",team);
        int ans=0;
        int flag=0;
        for(int i=0; i<n; i++)
        {if(n==1&&team[i]-‘0‘==0)
        {
            flag=1;
            break;
        }
            if(team[i]-‘0‘==0)
            {ans++;
            if(ans>=3)
            {
                flag=1;
                break;
            }
            if(ans==2)
            {
                if(i==1||i==n-1)
                {flag=1;
                break;
                }
            }
            }
            else
                if(team[i]-‘0‘==1)
            {
                ans=0;
                if(team[i-1]-‘0‘==1||team[i+1]-‘0‘==1)
                {
                    flag=1;
                    break;
                }
            }
        }
        if(flag)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}