Phone Number

一时失言乱红尘 2022-08-18 02:35 6阅读 0赞

## 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;  
    }