ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络练习赛 题目4 : Beautiful String

We say a string is beautiful if it has the equal amount of 3 or more continuous letters (in increasing order.)

Here are some example of valid beautiful strings: "abc", "cde", "aabbcc", "aaabbbccc".

Here are some example of invalid beautiful strings: "abd", "cba", "aabbc", "zab".

Given a string of alphabets containing only lowercase alphabets (a-z), output "YES" if the string contains a beautiful sub-string, otherwise output "NO".

输入

The first line contains an integer number between 1 and 10, indicating how many test cases are followed.

For each test case: First line is the number of letters in the string; Second line is the string. String length is less than 10MB.

输出

For each test case, output a single line "YES"/"NO" to tell if the string contains a beautiful sub-string.

提示

Huge input. Slow IO method such as Scanner in Java may get TLE.

样例输入

4
3
abc
4
aaab
6
abccde
3
abb

样例输出

YES
NO
YES
NO

 1 #include<iostream>
 2 #include<vector>
 3 #include<string>
 4 using  namespace std;
 5 
 6 int main() {
 7     int n;
 8     cin>>n;
 9     while( n-- ) {
10         vector<char> v;
11         int num;
12         cin>>num;
13         string str;
14         cin>>str;
15         vector<char> v1;
16         vector<int> v2;
17         v1.push_back(str[0]);
18         v2.push_back(1);
19         int flag = 1;
20     
21         for( int i = 1; i < str.size(); i++ ) {
22             if( str[i] == str[i-1] ) {
23                 int temp = *(v2.end()-1);
24                 temp++;
25                 v2.pop_back();
26                 v2.push_back(temp);
27                 if( v2[1] == v2[2] && v2[0] >= v2[1] && v1[2] - v1[1] == 1 && v1[1] - v1[0] == 1) {
28                         cout<<"YES"<<endl;
29                         flag = 0;
30                         break;
31                     }
32             }
33             else {
34                 if( v1.size() < 3 ) {
35                     v1.push_back(str[i]);
36                     v2.push_back(1);
37                     if( v2[1] == v2[2] && v2[0] >= v2[1] && v1[2] - v1[1] == 1 && v1[1] - v1[0] == 1) {
38                         cout<<"YES"<<endl;
39                         flag = 0;
40                         break;
41                     }
42                 }
43                 else {                    
44                         v1.erase(v1.begin());
45                         v2.erase(v2.begin());
46                         v1.push_back(str[i]);
47                         v2.push_back(1);
48                         if( v2[1] == v2[2] && v2[0] >= v2[1] && v1[2] - v1[1] == 1 && v1[1] - v1[0] == 1) {
49                         cout<<"YES"<<endl;
50                         flag = 0;
51                         break;
52                     }
53                     
54                 }
55             }
56         }    
57         if( flag ) 
58             cout<<"NO"<<endl;    
59     }
60     return 0;
61 }

 

posted on 2015-09-18 22:28  马衡  阅读(189)  评论(0)  编辑  收藏  举报

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