题目描述
We know thatif a phone number A is another phone number B’s prefix, B is not able to becalled. For an example, A is 123 while B is 12345, after pressing 123, we callA, and not able to call B.
Given N phone numbers, your task is to find whether there exits two numbers Aand B that A is B’sprefix.
输入
The input consists of several test cases.
The first line of input in each test case contains one integer N (0<N<1001),represent the number of phone numbers.
The next line contains N integers, describing the phonenumbers.
The last case is followed by a line containing one zero.
输出
For each test case, if there exits a phone number thatcannot be called, print “NO”, otherwise print “YES” instead.
示例输入
2
012
012345
2
12
012345
0
示例输出
NO
YES
/*********************
N个字符串中如果有一个是另外一个的前缀，则输出NO，否则输出YES。
O（n^2）的循环，
*********************/
Code:
#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
char str[1005][1005];
char s[1005];
int n,i,j,li,lj;
bool leaf;
while(scanf("%d",&n)&&n)
{
leaf = false;
li = lj = 0;
for(i = 0;i<n;i++)
cin>>str[i];
for(i = 0;i<n;i++)
{
li = strlen(str[i]);
for(j = 0;j<n;j++)
{
lj = strlen(str[j]);
if(i==j||li>lj)//自己不和自己比较，str[j] 长度比str[i]短的话肯定不是前缀。
continue;
strncpy(s,str[j],li);
s[li] = ' ';
if(strcmp(s,str[i])==0)// 发现前缀则跳出循环
{
leaf = true;
break;
}
}
if(leaf)
break;
}
if(leaf)
printf("NO
");
else
printf("YES
");
}
return 0;
}