codeforces 361 A - Mike and Cellphone-优快云博客

本文针对Codeforces竞赛中689A题进行了解析，该题涉及通过指纹记忆判断是否可能拨错电话号码的问题。通过模拟手指在数字键盘上的移动，算法检查是否存在另一种可能的数字组合，导致拨错电话。

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题目链接：http://codeforces.com/contest/689/problem/A

Description

While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital equal-sized keys, located in the following way:

$\text{[math]}$

Together with his old phone, he lost all his contacts and now he can only remember the way his fingers moved when he put some number in. One can formally consider finger movements as a sequence of vectors connecting centers of keys pressed consecutively to put in a number. For example, the finger movements for number "586" are the same as finger movements for number "253":

$\text{[math]}$ $\text{[math]}$

Mike has already put in a number by his "finger memory" and started calling it, so he is now worrying, can he be sure that he is calling the correct number? In other words, is there any other number, that has the same finger movements?

Input

The first line of the input contains the only integer n (1 ≤ n ≤ 9) — the number of digits in the phone number that Mike put in.

The second line contains the string consisting of n digits (characters from '0' to '9') representing the number that Mike put in.

Output

If there is no other phone number with the same finger movements and Mike can be sure he is calling the correct number, print "YES" (without quotes) in the only line.

Otherwise print "NO" (without quotes) in the first line.

Sample Input

Input

3 586

Output

NO

Input

2 09

Output

NO

Input

9 123456789

Output

YES

Input

3 911

Output

YES


题意：自己模拟了很久一直错，参考别人的循环列出可能按错键位的可能，如果有几个位置满足那个状态，那么他就有可能播错电话。

#include<iostream>
#include<cstdio>
using namespace std;
char a[12];
int n;
int xx()
{
    int a1,b,c,d;
    a1=1;
    b=1;
    c=1;
    d=1;
    for(int i=0; i<n; i++)
    {

        if(a[i]=='1'||a[i]=='4'||a[i]=='7'||a[i]=='0') a1=0;
        if(a[i]=='3'||a[i]=='6'||a[i]=='9'||a[i]=='0') b=0;
        if(a[i]=='1'||a[i]=='2'||a[i]=='3') c=0;
        if(a[i]=='7'||a[i]=='9'||a[i]=='0') d=0;
    }
    if(a1||b||c||d) return 0;
    else return 1;
}
int main()
{
    while(cin>>n)
    {
        cin>>a;
        if(xx()) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}