Beautiful Now
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
. 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?
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means
![](http://exam.upc.edu.cn/upload/image/20180806/20180806150711_45471.jpg)
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?
输入
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.
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.
输出
For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
样例输入
5
12 1
213 2
998244353 1
998244353 2
998244353 3
样例输出
12 21 123 321 298944353 998544323 238944359 998544332 233944859 998544332
暴搜,具体看注释:
#include <bits/stdc++.h> typedef long long ll; using namespace std; ll maxim=0,minum=1e10; int indexx=0; ll a[15]={0}; ll b[15]={0}; ll k; ll sum=0; void dfs1(int now,int cnt)//搜最大值 { if(cnt==k||now==indexx-1)//如果已交换,或者到了第N位 { ll temp=0; for(int i=0;i<indexx;i++) { temp=temp*10+a[i]; } maxim=max(maxim,temp); return; } dfs1(now+1,cnt);//不交换这一位 for(int i=now+1;i<indexx;i++) { if(a[now]<a[i])//贪心一下,当前这一位如果比下一位小,必然交换。不加貌似也能过 { swap(a[i],a[now]); dfs1(now+1,cnt+1); swap(a[i],a[now]); } } } void dfs2(int now,int cnt)//同理 { if(cnt==k||now==indexx-1) { ll temp=0; for(int i=0;i<indexx;i++) { temp=temp*10+a[i]; } minum=min(minum,temp); return; } dfs2(now+1,cnt); for(int i=now+1;i<indexx;i++) { if(now==0&&a[i]==0) continue; if(a[i]<a[now]) { swap(a[i],a[now]); dfs2(now+1,cnt+1); swap(a[i],a[now]); } } } int main() { int t; scanf("%d",&t); while(t--) { ll x,i; indexx=0; maxim=0; minum=1e10; memset(a,0,sizeof(a)); cin>>x>>k; while(x>0) { b[indexx++]=x%10; x/=10; } for(i=0;i<indexx;i++)//交换一下位置 { a[i]=b[indexx-1-i]; } //if(k>=index) k=index; dfs1(0,0); dfs2(0,0); cout<<minum<<" "<<maxim<<endl; } return 0; }