Codeforces Round #350 (Div. 2) F. Restore a Number 模拟构造题

F. Restore a Number

 

Vasya decided to pass a very large integer n to Kate. First, he wrote that number as a string, then he appended to the right integer k — the number of digits in n.

Magically, all the numbers were shuffled in arbitrary order while this note was passed to Kate. The only thing that Vasya remembers, is a non-empty substring of n (a substring of n is a sequence of consecutive digits of the number n).

Vasya knows that there may be more than one way to restore the number n. Your task is to find the smallest possible initial integer n. Note that decimal representation of number n contained no leading zeroes, except the case the integer n was equal to zero itself (in this case a single digit 0 was used).

Input

The first line of the input contains the string received by Kate. The number of digits in this string does not exceed 1 000 000.

The second line contains the substring of n which Vasya remembers. This string can contain leading zeroes.

It is guaranteed that the input data is correct, and the answer always exists.

Output

Print the smalles integer n which Vasya could pass to Kate.

Examples

input

003512
021

output

30021

 

题意：

　　

　　给你一个字符串a，b

　　字符串a是由  n字符串形式+n位数字符串形式 打乱的到的

　　且b是n的子串

　　现在问你能构成最小的n十多少

 

题解：

　　

　　观察到位数越小，n则会越小

　　我们枚举位数就好

　　然后就是一堆模拟

 

#include<bits/stdc++.h>
using namespace std;
const int N = 3e6+, M = 1e6+, mod = 1e9+, inf = 1e9+;
typedef long long ll;

int H[N];
vector<string > ans;
char a[N],sub[N];
int b[N],Sub,y[];
int pushdown(int len) {
    for(int i=;i<=;i++) H[i]-=y[i];
    int tmp = len,can = ;
        while(tmp) {
            if(H[tmp%]) H[tmp%] = H[tmp%]- ;
            else {can = ;H[tmp%] = H[tmp%]- ;}
            tmp/=;
    }
    int sum = ;for(int i=;i<=;i++) H[i]+=y[i];
    for(int i=;i<=;i++) sum+=H[i];

    if(sum!=len||!can) {
        tmp = len;
        while(tmp) {
            H[tmp%] = H[tmp%] + ;
            tmp/=;
        }
        return ;
    }
    else return ;
}

int main() {
    scanf("%s",a+);
    int L = strlen(a+);
    getchar();gets(sub+);
    Sub = strlen(sub+);
    for(int i=;i<=L;i++) H[a[i]-'']++;
    if(L==&&H[]==&&H[]==) {printf("0\n");return ;}
    for(int i=;i<=Sub;i++) y[sub[i]-'']++;
    for(int len = ;;len++) {
        if(!pushdown(len))continue;

           for(int i=;i<=;i++) H[i]-=y[i];
        int fir ,cnt = ;
        for(int i=;i<=;i++) {
            for(int j=;j<=H[i];j++) {
                b[++cnt] = i;
            }
        }
        //如果全部为0的情况
        if(b[cnt]==&&sub[]!='') {
            for(int i=;i<=Sub;i++) printf("%c",sub[i]);
            for(int i=;i<=cnt;i++) printf("%c",b[i]+'');
        }
        else {//找到Sub对应的位置就好了
           //     cout<<1<<endl;
            for(int i=;i<=cnt;i++) {
                if(b[i]) {
                    swap(b[],b[i]);
                    break;
                }
            }
            if(Sub==) {
                for(int i=;i<=cnt;i++) printf("%c",b[i]+'');
                return ;
            }

             int f = -;
                    for(int j=;j<=Sub;j++) {
                        if(sub[j]>sub[j-]) {
                            f=;break;
                        }
                        else if(sub[j]<sub[j-]) {f=;break;}
                        else continue;
                    }
            int yes = ;
            for(int i=;i<=cnt;) {
                if(i==&&sub[]!='') {
                    int l = ,can = ;
                    while(l<=Sub&&l<=cnt) {
                        if(b[l]+''<sub[l]) {can = ;break;}
                        else if(b[l]+''>sub[l]) {can = ;break;}
                        else {l++;}
                    }
                    if(can) {
                        cout<<sub+;
                        sort(b+,b+cnt+);
                        yes = ;
                    }
                     printf("%c",b[i++]+'');continue;
                }
                else if(i==){
                    printf("%c",b[i++]+'');
                    continue;
                }
                if(!yes) printf("%c",b[i++]+'');
                else {
                    int l = i;
                    int tmp = ;
                    while(l<=cnt&&b[l]+''<sub[tmp]) {
                        printf("%c",b[l++]+'');
                    }

                    if(f<=) {
                        printf("%s",sub+);
                    }
                    else {
                        while(l<=cnt&&b[l]+''==sub[tmp]) {
                         printf("%c",b[l++]+'');
                        }
                         printf("%s",sub+);
                    }
                    yes = ;
                    i = l;
                }
            }
            if(yes) cout<<sub+;
        }

       // cout<<" "<<len<<endl;
        printf("\n");return ;
    }
    return ;
}