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
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 22 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 23 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 24 //****************** 25 int abss(int a); 26 int lowbit(int n); 27 int Del_bit_1(int n); 28 int maxx(int a,int b); 29 int minn(int a,int b); 30 double fabss(double a); 31 void swapp(int &a,int &b); 32 clock_t __STRAT,__END; 33 double __TOTALTIME; 34 void _MS(){__STRAT=clock();} 35 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 36 //*********************** 37 #define rint register int 38 #define fo(a,b,c) for(rint a=b;a<=c;++a) 39 #define fr(a,b,c) for(rint a=b;a>=c;--a) 40 #define mem(a,b) memset(a,b,sizeof(a)) 41 #define pr printf 42 #define sc scanf 43 #define ls rt<<1 44 #define rs rt<<1|1 45 typedef pair<int,int> PII; 46 typedef vector<int> VI; 47 typedef long long ll; 48 const double E=2.718281828; 49 const double PI=acos(-1.0); 50 const ll INF=(1LL<<60); 51 const int inf=(1<<30); 52 const double ESP=1e-9; 53 const int mod=(int)1e9+7; 54 const int N=(int)1e6+10; 55 56 char a[50],b[50]; 57 ll get(char x[],int n) 58 { 59 ll temp=0; 60 for(int i=1;i<=n;++i) 61 temp*=10,temp+=x[i]-'0'; 62 return temp; 63 } 64 65 int main() 66 { 67 sc("%s%s",a+1,b+1); 68 int la=strlen(a+1); 69 int lb=strlen(b+1); 70 if(la<lb) 71 { 72 sort(a+1,a+1+la,greater<int>()); 73 pr("%s\n",a+1); 74 } 75 else if(la>lb) 76 { 77 pr("-1\n"); 78 } 79 else 80 { 81 sort(a+1,a+1+la); 82 ll Val=get(b,lb); 83 for(int i=1;i<=la;++i) 84 { 85 for(int j=i+1;j<=la;++j) 86 { 87 char temp[50]; 88 strcpy(temp+1,a+1); 89 swap(temp[i],temp[j]); 90 ll val=get(temp,la); 91 if(val<=Val) 92 swap(a[i],a[j]); 93 else 94 break; 95 } 96 } 97 ll ans=get(a,la); 98 pr("%lld\n",ans); 99 } 100 return 0; 101 } 102 103 /**************************************************************************************/ 104 105 int maxx(int a,int b) 106 { 107 return a>b?a:b; 108 } 109 110 void swapp(int &a,int &b) 111 { 112 a^=b^=a^=b; 113 } 114 115 int lowbit(int n) 116 { 117 return n&(-n); 118 } 119 120 int Del_bit_1(int n) 121 { 122 return n&(n-1); 123 } 124 125 int abss(int a) 126 { 127 return a>0?a:-a; 128 } 129 130 double fabss(double a) 131 { 132 return a>0?a:-a; 133 } 134 135 int minn(int a,int b) 136 { 137 return a<b?a:b; 138 }
