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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.
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.
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
#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; }
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