题目
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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
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
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.
样例
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Sample input:
3 Team000002 Rlsp0dfa Team000003 perfectpwd Team000001 R1spOdfa
Sample output:
2 Team000002 RLsp%dfa Team000001 R@spodfa
思路
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这题真的恶心，要看单复数。
如果不存在需要修改的密码，则输出There are n accounts and no account is modified。如果只有一个账户，就输出There is 1 account and no account is modified
完整代码
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#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int isok(string x){
int len = x.length();
for(int i = 0; i < len; i++){
if(x[i] == 'l') return 0;
if(x[i] == '1') return 0;
if(x[i] == '0') return 0;
if(x[i] == 'O') return 0;
}
return 1;
}
void deal(string x){
int len = x.length();
for(int i = 0; i < len; i++){
if(x[i] == 'l') x[i] = 'L';
if(x[i] == '1') x[i] = '@';
if(x[i] == '0') x[i] = '%';
if(x[i] == 'O') x[i] = 'o';
}
cout << x;
}
int main() {
int n;
int flag = 0;
int cnt = 0;
cin >> n;
int vis[1005];
string x1, x2;
string in1[1005];
string in2[1005];
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; i++){
cin >> x1 >> x2;
in1[i] = x1;
in2[i] = x2;
if(!isok(x2)){
flag = 1;
cnt ++;
vis[i] = 1;
}
}
if(!flag){
if(n == 1){
cout << "There is " << n << " account and no account is modified";
}else{
cout << "There are " << n << " accounts and no account is modified";
}
}else{
cout << cnt << endl;
for(int i = 0; i < 1005; i++){
if(vis[i]){
cout << in1[i] << " ";
deal(in2[i]);
cout << endl;
}
}
}
return 0;
}