SDNU 1135.Phone Number（水题）

桃扇骨 2021-12-22 14:09 190阅读 0赞

### 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][ACM 2010]

#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

[ACM 2010]: http://www.acmicpc.sdnu.edu.cn/problem/search?type=source&content=%E5%B1%B1%E4%B8%9C%E7%9C%81ACM%202010%E5%B9%B4%E7%AC%AC%E4%B8%80%E5%B1%8A%E7%9C%81%E8%B5%9B