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
題意
給了你n個4位數的pin code，你每次可以修改一個數的一個位置，問你最少修改多少次，可以使得每個數都不一樣。
題解
視頻題解 https://www.bilibili.com/video/av77514280/
數據范圍最多為10，所以每個pin code最多修改一個位置就可以了，我們首先給所有字符串標記出現過沒有，重復的就把他修改成沒有出現過的位置。
代碼
#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();
}
}