Codeforces Round #279 (Div. 2) E. Restoring Increasing Sequence

E. Restoring Increasing Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Peter wrote on the board a strictly increasing sequence of positive integers a1, a2, ..., an. Then Vasil replaced some digits in the numbers of this sequence by question marks. Thus, each question mark corresponds to exactly one lost digit.

Restore the the original sequence knowing digits remaining on the board.

Input

The first line of the input contains integer n (1 ≤ n ≤ 105) — the length of the sequence. Next n lines contain one element of the sequence each. Each element consists only of digits and question marks. No element starts from digit 0. Each element has length from 1 to 8 characters, inclusive.

Output

If the answer exists, print in the first line "YES" (without the quotes). Next n lines must contain the sequence of positive integers — a possible variant of Peter's sequence. The found sequence must be strictly increasing, it must be transformed from the given one by replacing each question mark by a single digit. All numbers on the resulting sequence must be written without leading zeroes. If there are multiple solutions, print any of them.

If there is no answer, print a single line "NO" (without the quotes).

Sample test(s)

input

3
?
18
1?

output

YES
1
18
19

input

2
??
?

output

NO

input

5
12224
12??5
12226
?0000
?00000

output

YES
12224
12225
12226
20000
100000

 思路: 1-n 处理，前面的已经确定了，处理的i 大于 i-1，而且取的尽量小

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 100010
using namespace std ;

char ans[maxn][10] ;
bool solve(char aa[],char b[])
{
    int n,m,i,j;
    char a[10] ;
    strcpy(a,aa) ;
    n = strlen(a) ;
    m = strlen(b) ;
    //cout<<n<<" "<<m<<endl;
    if(n<m) return false ;
    if(n>m)
    {
        if(a[0]=='?')a[0]='1';
        for( i = 1;i < n ;i++)if(a[i]=='?')
            a[i]='0';
        strcpy(aa,a) ;
        return true;
    }
    bool flag=false ;
    a[n]='1';
    b[n]='2';
    for(i = 0 ; i <= n ;i++)
    {
        if(a[i]=='?')
        {
            if(flag)a[i]='0';
            else a[i]=b[i] ;
        }
        else if(a[i]>b[i])flag=true;
        else if(a[i]<b[i])
        {
            if(flag) continue ;
            for( j = i ; j < n ;j++)
                if(a[j]=='?') a[j]='0' ;
            for(j = i ; j >= 0 ;j--)
            {
                if(aa[j]=='?')
                {
                    if(b[j]=='9')a[j]='0' ;
                    else
                    {
                        a[j]=b[j]+1;
                        a[n]='\0' ;
                        strcpy(aa,a) ;
                        return true;
                    }
                }
            }
            return false;
        }
    }
    a[n]='\0' ;
    strcpy(aa,a) ;
    return true;
}
int main()
{
    int i,n,m,j,k;
    char pre[10] ;
    while(scanf("%d",&n) != EOF)
    {
        for(i = 1 ; i <= n ;i++)
            scanf("%s",ans[i]) ;
        pre[0]='0';
        pre[1]='\0';
        for(i = 1 ; i <= n ;i++)
        {
            if(!solve(ans[i],pre)) break ;
            strcpy(pre,ans[i]);
        }
        if(i!=n+1)puts("NO") ;
        else
        {
            puts("YES") ;
            for( i = 1 ; i <= n ;i++)
               printf("%s\n",ans[i]) ;
        }
    }
    return 0;
}