甲级PAT 1035 Password （20 分）（字符串替换）

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Joyceyang_999

最新推荐文章于 2024-02-29 23:47:58 发布

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面对PAT竞赛，评委需生成随机密码，但1与l、0与O易混淆。本篇介绍如何使用C++代码检查并修改这些密码，确保清晰区分。

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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 %, lby 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

题目要求：

将账号对应的密码中易于混淆的字符替换，将1 (one) 替换为 @, 0 (zero) 替换为 %, l替换为 L, O(大写的o) 替换为 o

如果输入的所有账户信息中，密码中没有需要替换的。

如果账号数为1，输出There is 1 account and no account is modified

如果账号数为N，输出There are N accounts and no account is modified

解题思路：

条件判断，利用字符串里的replace方法。replace(需要替换的起始位置，替换长度，替换字符串）

完整代码：

#include<iostream>
#include<string>
using namespace std;

struct Account{
	string name;
	string result;
}; 

Account account[1010];
int M,num=0;

string convert(string s){
	int i;
	for(i=0;i<s.length();i++){
		if(s[i] == '1') s.replace(i,1,"@");
		else if(s[i] == '0') s.replace(i,1,"%");
		else if(s[i] == 'l') s.replace(i,1,"L");
		else if(s[i] == 'O') s.replace(i,1,"o");
	}
	return s;
}

int main(){
	int i;
	string s1,s2,s3;
	cin>>M;
	for(i=1;i<=M;i++){
		cin>>s1>>s2;
		s3 = convert(s2);
		if(s3!=s2){
			account[num].name = s1;
			account[num].result = s3;
			num++;
		}
	}
	if(num == 0){
		if(M == 1){
			printf("There is 1 account and no account is modified\n");
		}else{
			printf("There are %d accounts and no account is modified\n",M);
		}
		
	}else{
		cout<<num<<endl;
		for(i=0;i<num;i++){
			cout<<account[i].name<<" "<<account[i].result<<endl;
		}
	}
	return 0;
}