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  • codeforces 361 A


    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:

    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":

    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;
        }
    }
    View Code


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  • 原文地址:https://www.cnblogs.com/hfc-xx/p/5664783.html
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