To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N (≤1000), followed by N lines of accounts. Each account consists of a user name and a password, both are strings of no more than 10 characters with no space.

Output Specification:

For each test case, first print the number M of accounts that have been modified, then print in the following M lines the modified accounts info, that is, the user names and the corresponding modified passwords. The accounts must be printed in the same order as they are read in. If no account is modified, print in one line There are N accounts and no account is modified where N is the total number of accounts. However, if N is one, you must print There is 1 account and no account is modified instead.

Sample Input 1:

3
Team000002 Rlsp0dfa
Team000003 perfectpwd
Team000001 R1spOdfa

Sample Output 1:

2
Team000002 RLsp%dfa
Team000001 R@spodfa

Sample Input 2:

1
team110 abcdefg332

Sample Output 2:

There is 1 account and no account is modified

Sample Input 3:

2
team110 abcdefg222
team220 abcdefg333

Sample Output 3:

There are 2 accounts and no account is modified

题目翻译：

为了准备PAT考试，判定系统有时不得不给用户生成随机密码，但这样有个问题是总是会生成一些让人疑惑的密码，因为很难区分1和小写的l，0和大写的o。其中一个解决方案是将1用@替换，0用%替换，小写l用L替换，大写O用o替换。现在你的任务就是写一个程序来检查这些自动生成的密码，并且帮助判定系统来修改这些令人疑惑的密码。

输入规范：

每个输入文件包含一个测试用例，每个用例包含一个正整数N (≤1000),后面的N行为账户信息，每个账户信息由用户名和密码组成，都是不超过十位数且没有空格的字符串。

输出规范：

对于每个测试用例，首先第一行打印需要修改的账户个数M，紧接着的M行打印账号和修改后的密码，账户信息必须按照之前读入时的顺序读入。如果没有账户信息需要修改，则打印“There are N accounts and no account is modified”，但是如果N是一个，则就打印“There is 1 account and no account is modified”。

代码：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string str[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
int main(){
	string s;
	cin>>s;
	int sum=0;
	for(int i=0;i<s.length();i++)
		sum+=(s[i]-'0');
	string sum1=to_string(sum);
	for(int i=0;i<sum1.length();i++){
		cout<<str[sum1[i]-'0'];
		if(i<sum1.length()-1)
			cout<<" ";
	}
	return 0;
}