PAT-A1035 Password (密码替换)

A1035 Password (密码替换)

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结尾无空行

Sample Output 1:

2
Team000002 RLsp%dfa
Team000001 R@spodfa结尾无空行

Sample Input 2:

1
team110 abcdefg332结尾无空行

Sample Output 2:

There is 1 account and no account is modified

结尾无空行

Sample Input 3:

2
team110 abcdefg222
team220 abcdefg333结尾无空行

Sample Output 3:

There are 2 accounts and no account is modified

结尾无空行

题意：

给出数字N，循环输入N行学生的姓名和密码，学生的密码修改规则为，1 (one) by @, 0 (zero) by %, l by L, and O by o，要求输出更改密码的学生人数，并输出它们的姓名和更改后的密码，如果没有进行更改，则输出There is 1 account and no account is modified（注意复数形式！！！）

分析：

本题没什么难度，先用结构体接收学生信息，其中用flag来标识密码是否进行了更改，建立一个根据规则修改密码的方法change，如果返回值为true，则说明更改了密码，count++;最后进行输出。

代码如下：

#include<cstdio>
#include<string.h>
const int maxn=1010;
struct STU{
	char name[20];
	char pwd[20];
	bool flag;
}S[maxn];
bool flag;//不要在这里初始化
bool change(char pwd[]){
	flag=false;
	int len=strlen(pwd);
	for(int i=0;i<len;i++){
		if(pwd[i]=='1'){
			pwd[i]='@';  
			flag=true;
		}else if(pwd[i]=='0'){
			pwd[i]='%'; 
			flag=true;
		}else if(pwd[i]=='l'){
			pwd[i]='L'; 
			flag=true;
		}else if(pwd[i]=='O'){
			pwd[i]='o'; 
			flag=true;
		} 
	}
	return flag;
	
}
int main(){
	int n,count=0;
	//初始化
	for(int i=0;i<maxn;i++){
		S[i].flag=false;
	} 
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%s %s",&S[i].name,&S[i].pwd);
		if(change(S[i].pwd)){
			S[i].flag=true;
			count++;
		} 
	}
	if(count==0){
		if(n==1) printf("There is %d account and no account is modified",n);
		else printf("There are %d accounts and no account is modified",n);
	}else{
		printf("%d\n",count);
	for(int i=0;i<n;i++){
		if(S[i].flag){
			printf("%s %s\n",S[i].name,S[i].pwd);
		}
	}	
	}

	return 0;
}