Hiho for the 58th week. Beautiful String DP

Topic 1:beautiful String time limit:10000msSingle Point time limit:1000msMemory Limit:256MB

Describe

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

Here is some example of valid beautiful strings: "abc", "CDE", "AABBCC", "AAABBBCCC".

Here is 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".

Input

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

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

Output

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

Tips

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

Sample input

43abc4aaab6abccde3abb

Sample output

Yesnoyesno

 Test instructions, given a string, whether there is a contiguous substring, satisfies that the substring has more than 3 characters, is in ascending order and the number of each character is the same, such as a. Ab... BC...C where a b c is equal in number and is incremented. Because, when more than 3 times, always 3 characters are satisfied, so only consider is the case of 3, Num[i] that the first position, the back up to Num[i] characters are the same. If, there is such a x1...x2...x3...,x1x2x3 increment, and num[x2] = = Num[x3], num[x3] >= num[x1] (because such a string can exist AABBCCC, so long as, num[x3] >= x[1]) , str[x1] = = Str[i] + 1 &&  str[x2] = = Str[i] + 2&& num[x1] = = Num[i] && num[x2] >= num[i]& nbsp; the description exists. 

#define N 10000050#define M 100005#define maxn 205#define MOD 1000000000000000007int n,num[n],len;char str[n];int main () {            while (s (n)!=eof) {while (n--) {S (len);            Gets (str);            FI (len) {str[i] = GetChar ();            } num[len-1] = 1;                for (int i = len-2; i>=0;i--) {if (str[i] = = str[i+1]) num[i] = num[i+1] + 1;            else num[i] = 1;            } bool flag = FALSE;                FI (len) {int x1 = i + num[i],x2 = i + 2 * num[i],x3 = i + 3 * num[i];  if (x3 <= len && str[x1] = = Str[i] + 1 && str[x2] = = Str[i] + 2 && num[x1] = =                    Num[i] && num[x2] >= Num[i]) {flag = true;                Break            }} if (flag) {printf ("yes\n"); } else {printf ("no\n "); }}} return 0;}

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Hiho for the 58th week. Beautiful String DP