PAT甲级练习1035. Password (20)

1035. Password (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

3Team000002 Rlsp0dfaTeam000003 perfectpwdTeam000001 R1spOdfa

Sample Output 1:

2Team000002 RLsp%dfaTeam000001 R@spodfa

Sample Input 2:

1team110 abcdefg332

Sample Output 2:

There is 1 account and no account is modified

Sample Input 3:

2team110 abcdefg222team220 abcdefg333

Sample Output 3:

There are 2 accounts and no account is modified

应该还算简单的一题，注意题目的要求

一开始用char *而不是string来做，发现在struct中地址是连续的，直接输出会把name和pwd都输出，把flag插在两个char *中间会改善这种情况，但直接输出还是有些问题，所以还是用string来做好了，时间也充裕

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <map>#include <set>#include <string>#include <string.h>using namespace std;struct account{string name;bool flag;string pwd;}a[1000];int main(){bool iflag=false;int n, cnt=0;scanf("%d", &n);for(int i=0; i<n; i++){cin>>a[i].name>>a[i].pwd;a[i].flag = false;for(int j=0; j<a[i].pwd.length(); j++){if(a[i].pwd[j]=='1'){a[i].pwd[j] = '@';a[i].flag = true;iflag = true;}else if(a[i].pwd[j]=='0'){a[i].pwd[j] = '%';a[i].flag = true;iflag = true;}else if(a[i].pwd[j]=='l'){a[i].pwd[j] = 'L';a[i].flag = true;iflag = true;}else if(a[i].pwd[j]=='O'){a[i].pwd[j] = 'o';a[i].flag = true;iflag = true;}}if(iflag==true){cnt++;iflag = false;}}iflag = true;if(cnt!=0)printf("%d\n", cnt);for(int i=0; i<n; i++){if(a[i].flag==true){cout<<a[i].name<<" "<<a[i].pwd<<endl;iflag = false;}}if(iflag==true){if(n==1) printf("There is 1 account and no account is modified\n");else printf("There are %d accounts and no account is modified\n", n);}cin>>n;return 0;}