A<=B的前提下全排列A使答案尽量大

题意：http://codeforces.com/problemset/problem/915/C

举个例子：假使排好序后a字符串是123456，b是456456，按照上述方法遍历，
213456 ->312456->412356->(这是第一个字符的最大值，再往下变的话只能是5了，但不满足题意)—–>(这是就会发现原本最小的字符1在第二位了，这样的话就不会使你的遍历中间有漏项，因为第一位是当前最大的，第二位是当前最小的)—>再往下的同理，就不具体细推了
https://blog.csdn.net/yiqzq/article/details/79692352

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
int abss(int a);
int lowbit(int n);
int Del_bit_1(int n);
int maxx(int a,int b);
int minn(int a,int b);
double fabss(double a);
void swapp(int &a,int &b);
clock_t __STRAT,__END;
double __TOTALTIME;
void _MS(){__STRAT=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef long long ll;
const double E=2.718281828;
const double PI=acos(-1.0);
const ll INF=(1LL<<60);
const int inf=(1<<30);
const double ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;

char a[50],b[50];
ll get(char x[],int n)
{
    ll temp=0;
    for(int i=1;i<=n;++i)
        temp*=10,temp+=x[i]-'0';
    return temp;
}

int main()
{
    sc("%s%s",a+1,b+1);
    int la=strlen(a+1);
    int lb=strlen(b+1);
    if(la<lb)
    {
        sort(a+1,a+1+la,greater<int>());
        pr("%s
",a+1);
    }
    else if(la>lb)
    {
        pr("-1
");
    }
    else
    {
        sort(a+1,a+1+la);
        ll Val=get(b,lb);
        for(int i=1;i<=la;++i)
        {
            for(int j=i+1;j<=la;++j)
            {
                char temp[50];
                strcpy(temp+1,a+1);
                swap(temp[i],temp[j]);
                ll val=get(temp,la);
                if(val<=Val)
                    swap(a[i],a[j]);
                else
                    break;
            }
        }
        ll ans=get(a,la);
        pr("%lld
",ans);
    }
    return 0;
}

/**************************************************************************************/

int maxx(int a,int b)
{
    return a>b?a:b;
}

void swapp(int &a,int &b)
{
    a^=b^=a^=b;
}

int lowbit(int n)
{
    return n&(-n);
}

int Del_bit_1(int n)
{
    return n&(n-1);
}

int abss(int a)
{
    return a>0?a:-a;
}

double fabss(double a)
{
    return a>0?a:-a;
}

int minn(int a,int b)
{
    return a<b?a:b;
}