HDU 5311 Hidden String (elegant brute force), hdu5311

HDU 5311 Hidden String (elegant brute force), hdu5311

Hidden String Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission (s): 52 Accepted Submission (s): 25


Problem DescriptionToday is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string S Of length N . He wants to find three nonoverlapping substrings S [l1.. r1] , S [l2.. r2] , S [l3.. r3] That:

1. 1 ≤ l1 ≤ r1 <l2 ≤ r2 <l3 ≤ r3 ≤ n

2. The concatenation S [l1.. r1] , S [l2.. r2] , S [l3.. r3] Is "anniversary". Input There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤100) , Indicating the number of test cases. For each test case:
There's a line containing a string S (1 ≤ | s | ≤ 100) Consisting of lowercase English letters. output For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes ). sample Input

2annivddfdersewwefarynniversarya

Sample Output

YESNO

Source BestCoder 1st Anniversary ($)
Question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 5311

Question: In a string, we cannot find three consecutive intervals and splice them into anniversary.

Question Analysis: brute force, enumerative length of each segment

#include <cstdio>#include <cstring>char s[200], con[] = "anniversary";int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%s", s);        int len = strlen(s);        bool flag = false;        for(int i = 0; i <= 8; i++)        {            for(int j = i + 1; j <= 9; j++)            {                int k = 0;                while(k < len && strncmp(con, s + k, i + 1) != 0)                    k ++;                if(k == len)                     continue;                k += i + 1;                while(k < len && strncmp(con + i + 1, s + k, j - i) != 0)                    k ++;                if(k == len)                     continue;                k += j - i;                while(k < len && strncmp(con + j + 1, s + k, 10 - j) != 0)                    k ++;                if(k != len)                {                    flag = true;                    break;                }            }        }        if(flag)             puts("YES");        else             puts("NO");    }}

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