PAT甲级1035 Password

题目
1035 Password (20 分)
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

题目意思
是说1和 l 还有0和o 会让人分不清，所以定了一个规则，将1改为@，将l改为L，将0改为%，将O改为o，这样便区分开。输入一个n，表示有n组数据，检查n组数据里有几组需要按照规则更改，并将其按照规则改好。最后输出，有需要改的，就按照顺序输出。如都不需要更改则按照样例输出。
这里需要注意的是，不需要更改时，只有一个和有多个的输出有点不一样，account和accounts单复数的区别。

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>using namespace std;int main()
{int n,k=0,q=0,sum=0;scanf("%d",&n);char hx[1000][13],mz[1000][13];char a[13],b[13];for(int j=0;j<n;j++){scanf("%s %s",a,b);int flag=0;for(int i=0; i<strlen(b); i++){if(b[i]=='1'){b[i]='@';flag=1;}else if(b[i]=='l'){b[i]='L';flag=1;}else if(b[i]=='0'){b[i]='%';flag=1;}else if(b[i]=='O'){b[i]='o';flag=1;}}if(flag){sum++;strcpy(hx[k++],b);strcpy(mz[q++],a);}}if(sum){printf("%d\n",sum);for(int i=0;i<sum;i++){printf("%s %s\n",mz[i],hx[i]);}}else if(n==1)printf("There is 1 account and no account is modified\n");elseprintf("There are %d accounts and no account is modified\n",n);return 0;
}

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