1077 kuchiguse (20 points)

1077 Kuchiguse (20 branch )

The question

give n Group string , Find the maximum common suffix of this group of strings . If there is no output ”nai".

Ideas

First invert all strings , Change the public suffix to the public prefix .
Then iterate over all the strings , Find public prefix .
The input string contains spaces , Input char Type character array cannot be used scanf("%s"), It can be used cin.getline(str,m),str This is the string array ,m Subtract one from the maximum number of characters entered （ The terminator at the end of the string will occupy a ）. Also available string type , For input getline(cin,str).
string Type has string inversion function ,reverse(str,str+n) The length is n String inversion of .
and char You need to write an inversion function .
Pay attention to differences string and char Different input and output modes .

Code

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

using namespace std;

char str[205][258];

void reverse(int n)
{
	int len;
	for(int i=n-1;i>=0;i--){
		len=strlen(str[i]);
		for(int j=0;j<len;j++){
			str[i+1][j]=str[i][len-1-j];
		}
	}
}

void cmp(int n,int m)
{
	int count=0;
	bool flag=0,temp=0;
	for(int i=0;i<m;i++){
		for(int j=2;j<=n;j++){
			if(str[j][i]!=str[1][i]){
				flag=1;
				break;
			}
		}
		if(flag){
			break;
		}
		count++;
	}
	if(count==0){
		printf("nai");
	}else{
		for(int i=count-1;i>=0;i--){
			printf("%c",str[1][i]);
		}
	}
}

int main()
{
	int n,m=256;
	
	scanf("%d\n",&n);
	for(int i=0;i<n;i++){
		cin.getline(str[i],257);
		if(m>strlen(str[i])){
			m=strlen(str[i]);
		} 
	}
	reverse(n);
	cmp(n,m);
	return 0;
}