Codeforces Round #603 (Div. 2) B. PIN Codes 水题

B. PIN Codes

A PIN code is a string that consists of exactly 4 digits. Examples of possible PIN codes: 7013, 0000 and 0990. Please note that the PIN code can begin with any digit, even with 0.

Polycarp has n (2≤n≤10) bank cards, the PIN code of the i-th card is pi.

Polycarp has recently read a recommendation that it is better to set different PIN codes on different cards. Thus he wants to change the minimal number of digits in the PIN codes of his cards so that all n codes would become different.

Formally, in one step, Polycarp picks i-th card (1≤i≤n), then in its PIN code pi selects one position (from 1 to 4), and changes the digit in this position to any other. He needs to change the minimum number of digits so that all PIN codes become different.

Polycarp quickly solved this problem. Can you solve it?

Input

The first line contains integer t (1≤t≤100) — the number of test cases in the input. Then test cases follow.

The first line of each of t test sets contains a single integer n (2≤n≤10) — the number of Polycarp's bank cards. The next n lines contain the PIN codes p1,p2,…,pn — one per line. The length of each of them is 4. All PIN codes consist of digits only.

Output

Print the answers to t test sets. The answer to each set should consist of a n+1 lines

In the first line print k — the least number of changes to make all PIN codes different. In the next n lines output the changed PIN codes in the order corresponding to their appearance in the input. If there are several optimal answers, print any of them.

Example

input

3

2

1234

0600

2

1337

1337

4

3139

3139

3139

3139

output

0

1234

0600

1

1337

1237

3

3139

3138

3939

6139

题意

给了你n个4位数的pin code，你每次可以修改一个数的一个位置，问你最少修改多少次，可以使得每个数都不一样。

题解

视频题解 https://www.bilibili.com/video/av77514280/

数据范围最多为10，所以每个pin code最多修改一个位置就可以了，我们首先给所有字符串标记出现过没有，重复的就把他修改成没有出现过的位置。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 15;
string s[15];
int n,ans;
map<string,int>H;
void change(int x){
	H[s[x]]--;
	for(int i=0;i<s[x].size();i++){
		char ori = s[x][i];
		for(int j=0;j<10;j++){
			char c = '0'+j;
			s[x][i]=c;
			if(!H[s[x]]){
				ans++;
				H[s[x]]=1;
				return;
			}
		}
		s[x][i]=ori;
	}
}
void solve(){
	H.clear();
	ans = 0;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>s[i];
		H[s[i]]++;
	}
	//sort(s,s+n);
	for(int i=0;i<n;i++){
		if(H[s[i]]>1){
			change(i);
		}
	}
	cout<<ans<<endl;
	for(int i=0;i<n;i++){
		cout<<s[i]<<endl;
	}
}
int main(){
	int t;
	cin>>t;
	while(t--){
		solve();
	}
}