Permute Digits

最新推荐文章于 2022-10-27 17:29:32 发布

非人类浩

最新推荐文章于 2022-10-27 17:29:32 发布

阅读量360

点赞数

CC 4.0 BY-SA版权

分类专栏： 17寒假蓝桥文章标签：回溯

本文链接：https://blog.youkuaiyun.com/C___J___H/article/details/79175272

17寒假蓝桥专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个名为D-PermuteDigits的算法问题，该问题要求通过改变一个给定正整数a的数字顺序来构造一个不超过另一个给定正整数b的最大可能数值。文章详细解释了算法的具体实现过程，包括输入输出格式、示例和核心代码逻辑。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

D - Permute Digits
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.

Example
Input
123
222
Output
213
Input
3921
10000
Output
9321
Input
4940
5000
Output
4940

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int k = 0;
int main(){
    char a[20],b[20];
    int ans[20],cnt[10],flag,flag1;
    memset(cnt,0,sizeof(cnt));
    flag = -2;
    flag1 = 0;
    cin>>a>>b;
    for(int i = 0;i < strlen(a);i++)
        cnt[a[i]-'0']++;
    ///位数比b小，直接逆序输出 
    if(strlen(a)<strlen(b)){
        for(int i = 9;i>=0;i--){
            while(cnt[i]-- != 0) ans[k++] = i;
        }
    }
    ///位数相等的时候 
    else{
        while(k<strlen(a)){
            ///判断是否需要回溯 
            if(flag == 1){
                k--;
                cnt[ans[k]]++;
                for(int i = b[k]-'0'-1;i>=0;i--){
                    if(cnt[i]!=0){
                        ans[k] = i;
                        cnt[i]--;
                        flag = -1;
                        break;
                    }
                }
                ///如果回溯，此时一定比b该位的数小，后面直接逆序输出 
                if(flag == -1) break;
                if(flag == 1) continue;         
            }
            ///如果i位和b该位相等继续循环 
            if(cnt[b[k]-'0']!=0){
                cnt[b[k]-'0']--;
                ans[k] = b[k]-'0';
            }
            ///如果i位比b该位小，打破循环，后面逆序输出 
            else{
                flag1 = 0;
                for(int i = (b[k]-'0')-1;i>=0;i--){
                    if(cnt[i]!=0){
                        ans[k] = i;
                        cnt[i]--;
                        flag1 = flag = -1;
                        break;
                    }
                }
                if(flag1 != -1) flag = 1;
            }
            if(flag == -1) break;
            if(flag==1) continue;
            k++;
        }

    }
    ///中途跳出来循环 剩下的逆序输出 
    if(flag == -1){
        for(int i = 9;i >= 0;i--)
            while(cnt[i]!=0){
                ans[++k] = i;
                cnt[i]--;   
            }
        for(int i = 0;i < strlen(a);i++)
            cout<<ans[i];
            cout<<endl; 
    }
    ///中途为打破循环 等于b或者位数比b小 
    else{
        for(int i = 0;i < strlen(a);i++)
            cout<<ans[i];
            cout<<endl; 
    }
    return 0;
}