1035 password (20 points)

1035 Password (20 branch )

The question

Give a set of member information , Include member name and password , If there is... In the password ‘1’ Change to ‘@’, Yes ‘0’ Change to ‘%’, Yes ‘l’（ A lowercase letter L） Change to ’L‘, Yes ’O‘ Change to ’o’.
Output the number of people to change the password , And the modified information of these people （ name Modified password ）
If no one's password needs to be changed , The output “There are N accounts and no account is modified”,N Is the total number of restructured members , If N by 1, The output of “There is 1 account and no account is modified”

Ideas

Create a member information structure , Include name and password , And a modification bit . Check whether the password needs to be modified and modify , And set the modification of the member to 1, It is convenient for judgment when outputting the information of modifying members later .
Finally, judge the total number of people and the number of people to be modified , Output the information of the member to be modified .
Be careful N by 1 Different output at .

Code

#include<stdio.h>
#include<string.h>

typedef struct node{
	char name[20],password[20];
	bool modify=0;
}node;

node member[2007];

void print(int n,int count)
{
	if(count==0){
		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);
		}
	}else{
		printf("%d\n",count);
		for(int i=0;i<n;i++){
			if(member[i].modify==1){
				printf("%s %s\n",member[i].name,member[i].password);
			}
		}
	}
}

int main()
{
	
	int n,count=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		getchar();
		scanf("%s %s",member[i].name,member[i].password);
		for(int j=0;j<strlen(member[i].password);j++){
			if(member[i].password[j]=='1'){
				member[i].modify=1;
				member[i].password[j]='@';
			}
			if(member[i].password[j]=='0'){
				member[i].modify=1;
				member[i].password[j]='%';
			}
			if(member[i].password[j]=='l'){
				member[i].modify=1;
				member[i].password[j]='L';
			}
			if(member[i].password[j]=='O'){
				member[i].modify=1;
				member[i].password[j]='o';
			}
		}
		if(member[i].modify==1){
			count++;
		}
	}
	print(n,count);
	return 0;
}