Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers:
Emergency 911Alice 97 625 999Bob 91 12 54 26
In this case, it's not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob's phone number. So this list would not be consistent.
输入描述:The first line of input gives a single integer, 1≤t≤401 leq t leq 401≤t≤40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1≤n≤100001 leq n leq 100001≤n≤10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.输出描述:
For each test case, output "YES" if the list is consistent, or "NO" otherwise.
示例1
输入2 3 911 97625999 91125426 5 113 12340 123440 12345 98346输出
NO YES
题目分析:
在一个数组中,如果一个数是另一个数的前缀和,输出NO,否则输出YES。
#include
using namespace std;
int main()
{
int t;
cin>>t;
string s[100003];
while(t--)
{
int n;
cin>>n;
for(int i=0;i
sort(s,s+n);
int q=0;
for(int i=0;i
string a,b;
a=s[i],b=s[i+1];
if(b.find(a)==0)//在b中找a
q=1;
}
if(q)
cout<<"NO"<
cout<<"YES"<
}
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