CF915C Permute Digits 字符串贪心-白红宇

CF915C Permute Digits 字符串贪心

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发布时间：2019-06-20

本文共 2000 字，大约阅读时间需要 6 分钟。

You are given two positive integer numbers a and b. Permute (change order) of the digits of a to construct maximal number not exceeding b. No number in input and/or output can start with the digit 0.

It is allowed to leave a as it is.

Input

The first line contains integer a (1 ≤ a ≤ 10¹⁸). The second line contains integer b(1 ≤ b ≤ 10¹⁸). Numbers don't have leading zeroes. It is guaranteed that answer exists.

Output

Print the maximum possible number that is a permutation of digits of a and is not greater than b. The answer can't have any leading zeroes. It is guaranteed that the answer exists.

The number in the output should have exactly the same length as number a. It should be a permutation of digits of a.

Examples

Input

123 222

Output

Input

3921 10000

Output

Input

4940 5000

Output

4940 题意：给你两个小于10^18的数a,b，a的每一位的数可以随意排列，求所得到的小于b的最大a 分析：每次暴力判断每一位数放前面时后面取最大是否大于b，如果不大于则当前位取这个数是最佳 AC代码：

#include #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   
                    #include 
                    
                     #define ls (r<<1)#define rs (r<<1|1)#define debug(a) cout << #a << " " << a << endlusing namespace std;typedef long long ll;const ll maxn = 1e4+10;const ll mod = 1000000007;const double pi = acos(-1.0);const double eps = 1e-8;bool cmp( char p, char q ) { return p > q;}int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); string s1, s2; while( cin >> s1 >> s2 ) { ll len1 = s1.length(), len2 = s2.length(); sort(s1.begin(),s1.end()); if( len1 < len2 ) { reverse(s1.begin(),s1.end()); cout << s1 << endl; continue; } for( ll i = 0; i < len1; i ++ ) { for( ll j = len1-1; j > i; j -- ) { string t = s1; swap(s1[i],s1[j]); sort(s1.begin()+i+1,s1.end()); if( s1 > s2 ) { s1 = t; //每次循环确定一个当前所能取到的最小值中的最大值数字 } else { break; } } //debug(s1); } cout << s1 << endl; } return 0;}

转载于:https://www.cnblogs.com/l609929321/p/9720806.html

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