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<<);
 const int inf=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e6+;

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

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

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

 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-);
 }

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

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

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