ZOJ 3697 Bad-written Number

Description

A little kid loves to write digits in LC display style, but he would like to put two consecutive digits very close to each other. Sometimes, that makes the number he writes ambiguous.

In LC display style, each digit is written as below:

As the table shown above, in LC display style, each digit is written in 3 rows and 3 columns. The little kid always overlaps the last column of the previous digit and the first column of the next digit.

Please help his parents to count the number of different ways to translate the weird 'image'.

Input

There are multiple test cases. The first line of input contains an integer T (T ≤ 1500) indicating the number of test cases. Then T test cases follow.

The first line of each case contains one integer: n (n ≤ 10⁴) -- the number of digits that the little kid writes.

Then each of the following 3 lines contains exactly 2n+1 characters -- represents the number written by the kid.

Output For each test case, print a single number -- the number of ways to express the bad-written 'image' module 10  ⁹+7 (which is equivalent to print 

answer % 1000000007, where 

% is the module operator in all major programming languages).

Sample Input

6
1
   
|  
|  
1
 _ 
|_|
|_ 
1

  |
  |
1
 _
|_|
|_|
2
   _ 
  |_|
  |_|
2
 _ _ 
 _|_ 
|_ _|

Sample Output

0
0
1
1
3
1

动态规划，判断有几种可行方案，比较麻烦的是判断数字相连的这部分内容。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<math.h>
using namespace std;
const int N=1e5+10;
const int mod=1e9+7;
int T,n;
char s[3][N];
long long dp[N][10];

int get(int x,int y)
{
    x=(x-1)*2;
    if (s[1][x]==' '&&(y==0||y==4||y==5||y==6||y==8||y==9)) return 0;
    if (s[2][x]==' '&&(y==0||y==2||y==6||y==8)) return 0;
    x++;
    if (s[0][x]==' '&& y!=1&&y!=4) return 0;
    if (s[0][x]=='_'&& (y==1||y==4)) return 0;
    if (s[1][x]==' '&&y!=1&&y!=0&&y!=7) return 0;
    if (s[1][x]=='_'&&(y==1||y==0||y==7)) return 0;
    if (s[2][x]==' '&&y!=1&&y!=4&&y!=7) return 0;
    if (s[2][x]=='_'&&(y==1||y==4||y==7)) return 0;
    x++;
    if (s[1][x]==' '&&(y!=5&&y!=6)) return 0;
    if (s[2][x]==' '&&y!=2) return 0;
    return 1;
}

int check(int x,int y,int g)
{
    int a = (x!=5&&x!=6&&x!=-1)||(y!=1&&y!=2&&y!=3&&y!=7&&y!=-1);
    int b = (x!=2&&x!=-1)||(y!=1&&y!=3&&y!=4&&y!=5&&y!=7&&y!=9&&y!=-1);
    g=(g-1)*2;
    if (s[1][g]=='|' && !a) return 0;
    if (s[1][g]==' ' && a) return 0;
    if (s[2][g]=='|' && !b) return 0;
    if (s[2][g]==' ' && b) return 0;
    return 1;
}

int main()
{
    for (scanf("%d",&T);T--;)
    {
        scanf("%d",&n); getchar();
        gets(s[0]); gets(s[1]); gets(s[2]);
        for (int j=0;j<3;j++)
        {
            int len=strlen(s[j]);
            for (int i=0;i<=2*n;i++)
            {
                if (i>=len) s[j][i]=' ';
            }
        }
        //puts(s[0]+1); puts(s[1]+1); puts(s[2]+1);
        for (int i=0;i<=9;i++)
        {
            dp[1][i]=get(1,i)&&check(-1,i,1);
        }
        for (int i=2;i<=n;i++)
        {
            for (int j=0;j<10;j++)
            {
                dp[i][j]=0;
                if (!get(i,j)) continue;
                for (int k=0;k<10;k++)
                {
                    (dp[i][j]+=dp[i-1][k]*check(k,j,i))%=mod;
                }
            }
        }
        long long ans=0;
        for (int i=0;i<10;i++)
        {
            (ans+=dp[n][i]*check(i,-1,n+1))%=mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}