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
2
012
012345
2
12
012345
0
Sample Output
NO
YES
Source
山东省ACM 2010年第一届省赛
#include <cstdio>
#include <iostream>
#include <cmath>
#include <
string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
int n;
string s[
1000+
8];
int main()
{
while(~scanf(
"%d", &n) && n !=
0)
{
for(
int i =
0; i<n; i++
)
cin>>
s[i];
bool flag =
1;
for(
int i =
0; i<n; i++
)
{
for(
int j = i+
1; j<n; j++
)
{
int len1 = s[i].size(), len2 =
s[j].size();
for(
int k =
0; k<min(len1, len2); k++
)
{
if(s[i][k] ==
s[j][k])
{
flag =
0;
break;
}
}
if(!flag)
break;
}
if(!flag)
break;
}
if(!flag)printf(
"NO\n");
else printf(
"YES\n");
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/10985742.html
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