CF915C Permute Digits 字符串贪心-优快云博客

本文探讨了在给定两个正整数a和b的情况下，如何通过改变a的数字顺序来构建一个不超过b的最大可能数字。文章提供了一个详细的AC代码示例，展示了如何通过比较和置换数字来达到目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using 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