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();
}
}

最新文章

  1. 通用数据库操作类,前端easyui-datagrid,form
  2. Qt 窗口等设置
  3. BZOJ 2818GCD
  4. javascript DOM对象转jquery对象
  5. WEB服务器1--开篇
  6. MySql命令——show,分页,正则表达式
  7. DataGridView常用功能
  8. mysql密码忘记如何处理
  9. IIS8中添加WCF支持几种方法小结[图文]
  10. El表达式取map值
  11. 简单的一句sql
  12. JSP的内置对象以及作用域。
  13. 基于Twitter的Snowflake算法实现分布式高效有序ID生产黑科技(无懈可击)
  14. [2019BUAA人工智能实战_陈泽寅]第1次个人作业
  15. java课程课后作业190425之一维数组最大子数组(界面实现)
  16. mfc简单框架的分析和原理记录
  17. Linux 线程编程2.0——线程同步-互斥锁
  18. Java中终止线程的三种方法
  19. 回文日期(NOIP2016)
  20. 【BZOJ1304】[CQOI2009]叶子的染色(动态规划)

热门文章

  1. 【Puppeteer】puppeteer安装/常用的方法以及一个小栗子(Youtube油管自动评论)
  2. 通过传XML格式导入到ORACLE的销售订单
  3. 百度大脑UNIT3.0详解之语音语义一体化方案
  4. solo升级以及自动化更新的方法
  5. 使用Python轻松批量压缩图片
  6. linux下安装make工具
  7. 22(7).模型融合---CatBoost
  8. 搭建Harbor
  9. acwing 853. 有边数限制的最短路 模板
  10. vue定义全局date过滤器(自定义JS文件模块和Moment.js库)