Min Number FZU - 2111(模拟)

本文介绍了一个关于如何通过交换数字中的两个位来最小化一个非负整数的问题,并提供了一段C++代码实现。该问题的目标是在不超过指定次数的情况下找到最小可能的10进制表示形式,不允许出现前导零。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Min Number

FZU - 2111

Now you are given one non-negative integer n in 10-base notation, it will only contain digits ('0'-'9'). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].

For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.

Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?

Please note that in this problem, leading zero is not allowed!


Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.

Output
For each test case, output the minimum number we can get after no more than M operations.
Sample Input
3
9012 0
9012 1
9012 2
Sample Output
9012
1092
1029

模拟

code:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
char str[1005],s[1005];
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int m;
        scanf("%s%d",str,&m);
        strcpy(s,str);//重新将str存一遍
        int n = strlen(s);
        sort(s,s+n);//给新的字符串从小到大排序
        for(int i = 0; i < n; i++){
            if(s[i] != '0'){//把前导零和最小的非零数交换,得到满足题意的最小数字
                swap(s[i],s[0]);
                break;
            }
        }
        for(int i = 0; i < n; i++){
            if(m == 0) break;
            if(s[i] == str[i]) continue;//如果和最小情况数字相同就不需要换
            char tmp = '9';
            int x = i+1;
            for(int j = i+1; j < n; j++){
                if(i == 0 && str[j] == '0') continue;//如果前面位置是第一个位置并且后面位置数字是0不能换,因为换了会出现前导零
                if(str[j] <= tmp){//找出最后面最小的数字
                    x = j;
                    tmp = str[j];
                }
            }
            swap(str[i],str[x]);//交换
            m--;
        }
        printf("%s\n",str);
    }
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值