Uva 省赛傻逼题 C

呵呵居然怕字符串不做，有好有坏！

Problem C

Time limit: 1.000 seconds 

 

Character Recognition?

Write a program that recognizes characters. Don't worry, because you only need to recognize three digits: 1, 2 and 3. Here they are:

.*.  ***  ***
.*.  ..*  ..*
.*.  ***  ***
.*.  *..  ..*
.*.  ***  ***

Input

The input contains only one test case, consisting of 6 lines. The first line contains n, the number of characters to recognize (1<=n<=10). Each of the next 5 lines contains 4n characters. Each character contains exactly 5 rows and 3 columns of characters followed by an empty column (filled with '.').

Output

The output should contain exactly one line, the recognized digits in one line.

Sample Input

3
.*..***.***.
.*....*...*.
.*..***.***.
.*..*.....*.
.*..***.***.

Output for the Sample Input

123

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The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan

 

#include <map>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1111
#define ll long long
int n,m;
int ans[maxn];
char s[maxn];
int main(){
    while(~scanf("%d",&n)){
        for(int i=0;i<3;i++)scanf("%s",s);
        scanf("%s",s);
        for(int i=0;i<4*n;i+=4){
            if(s[i+1]=='*')ans[i/4]=1;
            if(s[i]=='*')ans[i/4]=2;
            if(s[i+2]=='*')ans[i/4]=3;
        }
        scanf("%s",s);
        for(int i=0;i<n;i++)printf("%d",ans[i]);
        printf("\n");
    }
    return 0;
}