Phone Number
Time Limit: 1000MS Memory limit: 65536K
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.
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.
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.
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
#include<iostream>
#include<string>
using namespace std;
string a[1005],k; int b[1005];
int main()
{
int n,i,j,p,s;
while(cin>>n&&n)
{
s=0;
for(i=0;i<n;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++)
//cout<<a[i]<<endl;
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;
}
// for(i=0;i<n-1;i++)
// if(a[i]==a[i+1])
// s=2;
if(s==1)
cout<<"NO"<<endl;
else
if(s==0)
cout<<"YES"<<endl; } return 0;
}