A

Description

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.

Input

 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.

Output

 For each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.

Sample Input

20120123452120123450

Sample Output

NOYES

 

解题思路：

利用字符串数组，对于每个字符串查找它能否成为其它字符串的前缀，

细节处理：

若a是b的前缀则b比a长，对b从前往后取与a等长的字符串与a进行比较，如果只有一个字符串直接输出YES

代码：

#include <iostream>#include<string>#include<cstring>using namespace std;int main(){    int n,i,j,lena,lenb;    string a[1009];    while(cin>>n)    {        if(n==0) break;        for(i=1;i<=n;i++)        cin>>a[i];        if(n==1) {cout<<"YES"<<endl; continue;}        int flag=0;        for(i=1;i<=n;i++)        {            lena=a[i].length();            for(j=1;j<=n;j++)            {                lenb=a[j].length();                if(lenb>=lena&&i!=j)                {                    string t;                    for(int r=0;r<lena;r++)                    {t+=a[j][r];}                    if(t==a[i])                    {flag=1; break;}                }            }        }        if(flag) cout<<"NO"<<endl;        else cout<<"YES"<<endl;    }    return 0;}