CodeForces 689A Mike and Cellphone （模拟+水题）

Mike and Cellphone

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#problem/E

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：

判断是否有路径形状一致的按键序列；

题解：

正确建坐标系来量化各按键间的转移过程，暴力求解；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
#include<list>
#define LL long long
#define maxn 210000
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;

int n;
struct node{
    int x,y;
};
node pos[] = {{2,4},{1,1},{2,1},{3,1},{1,2},{2,2},{3,2},{1,3},{2,3},{3,3}};
int path[15];

bool is_ok(int x,int y) {
    if(x==1&&y==4 || x==3&&y==4) return 0;
    return x>=1&&x<=3&&y>=1&&y<=4;
}

int main(int argc, char const *argv[])
{
   //IN;

    while(scanf("%d",&n) != EOF)
    {
        getchar();
        for(int i=1; i<=n; i++){
            int c = getchar();
            path[i] = c-'0';
        }

        int flag = 1;
        for(int i=0; i<=9; i++){
            if(i == path[1]) continue;
            flag = 1;
            int curx = pos[i].x;
            int cury = pos[i].y;
            for(int j=2; j<=n; j++){
                int dx = pos[path[j]].x - pos[path[j-1]].x;
                int dy = pos[path[j]].y - pos[path[j-1]].y;
                curx += dx; cury += dy;
                if(!is_ok(curx,cury)) {flag=0;break;}
            }
            if(flag) {puts("NO");break;}
        }

        if(!flag) puts("YES");
    }

    return 0;
}