Phone Number

Phone Number

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

题目描述

We know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345,
after pressing 123, we call A, and not able to call B.

Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.

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 phone numbers.

The last case is followed by a line containing one zero.

输出

For each test case, if there exits a phone number that cannot 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

提示

来源

2010年山东省第一届ACM大学生程序设计竞赛

示例程序

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
/*struct node
{
int num;
char name[10000];
}a[1010],b;
int cmp(const void *a,const void *b)
{
return (*(struct node *)a).num>(*(struct node *)b).num?1:-1;
}*/
string a[1005],k;
int b[1005];
int main()
{
int n,i,j,p,s;
while(scanf("%d",&n)&&n)
{
s=0;
for(i=0;i<n;i++)
/*{
scanf("%s",a[i].name);
a[i].num=strlen(a[i].name);
}
qsort(a,n,sizeof(a[0]),cmp);*/
//gets(a[i]);
//scanf("%s",a[i]);
cin>>a[i];
for(i=0;i<n-1;i++)
for(j=0;j<n-1-i;j++)
{
if(a[j]>a[j+1])
{
k=a[j];
a[j]=a[j+1];
a[j+1]=k;
}
}
for(i=0;i<n;i++)
b[i]=a[i].size();
for(i=0;i<n-1;i++)
{
p=b[i]<b[i+1]?b[i]:b[i+1];
for(j=0;j<p;j++)
if(a[i][j]!=a[i+1][j])
break;
else if(j==p-1)
s=1;
}
if(s==1)
printf("NO\n");
else
if(s==0)
printf("YES\n");
}
return 0;
}