Problem Description
Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m−i. In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?
Input
The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.
Output
For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
Sample Input
5 12 1 213 2 998244353 1 998244353 2 998244353 3
Sample Output
12 21 123 321 298944353 998544323 238944359 998544332 233944859 998544332
Source
2018 Multi-University Training Contest 5
题意:给定一个数,和一个最多交换次数k,问在不超过k次操作的情况,问可以得到的最大值和最小值是多少?
思路:保存下标,全排列枚举下标。
难点:对于每次的排列,如何确定它的交换次数?看循环结的个数就行!
具体看代码:
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#include <stack>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define Lson l,m,rt<<1
#define Rson m+1,r,rt<<1|1
const int maxn = 1e5+10;
const int mod=1<<30;
#define inf 0x3f3f3f3f
int c[maxn];
char s[20];
int n;
bool vis[20];
int k;
bool check()
{
memset(vis,0,sizeof(vis));
int tmp=0;
for(int i=0;i<n;i++)
{
if(vis[i]) continue;
int x=0;
while(vis[i]==0)
{
x++;
vis[i]=1;
i=c[i];
}
tmp+=x-1;
if(tmp>k) return 0;
}
return tmp<=k;
}
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int T;
cin>>T;
while(T--)
{
scanf("%s%d",s,&k);
int Min=1e9+7;
int Max=0;
n=strlen(s);
for(int i=0;i<n;i++)
{
c[i]=i;
}
do
{
if(s[c[0]]!='0'&&check())
{
ll ans=0;
for(int i=0;i<n;i++)
{
ans=ans*10+(s[c[i]]-'0');
}
if(ans>Max)
{
Max=ans;
}
if(ans<Min)
{
Min=ans;
}
}
}while(next_permutation(c,c+n));
cout<<Min<<" "<<Max<<endl;
}
return 0;
}