Codechef August lunchtime 2014

Question

A string composed of two types of characters, A and B, each time you can select any length of the ECHO sub-sequence (not necessarily consecutive) and delete. It takes at least a few times to empty the entire string.

Idea: if it is a text return string, it only needs one time; otherwise, it needs two times (select all a for the first time and select all B for the second time ).

Accepted code:

 1 def is_palidrome(s): 2       n = len(s); 3       for i in xrange(n / 2): 4           if s[i] != s[n - i - 1]: 5             return False; 6     return True; 7  8 if __name__ == ‘__main__‘: 9     T = int(input())10     while T:11         T -= 1;12         H = raw_input();13         print 1 if is_palidrome(H) else 2;

Question B

Knowledge point: the formula ind = (A * ind + B) % m is given, and the length of the cyclic section is obtained. From the formula, we can see that the cyclic section does not exceed M.

In addition, the accuracy of this question is also a pitfall, which can be manually output: ". 5" and ". 0 ".

Accepted code:

1 /************************************** * *********************************** 2> File Name: wall. CPP 3> author: stomach_ache 4> mail: [email protected] 5> created time:, Sunday, August 31, 2014 6> propose: 7 *************************************** * *******************************/8 9 # include <cmath> 10 # include <string> 11 # include <cstdio> 12 # include <vector> 13 # include <iomanip> 14 # include <Fstream> 15 # include <cstring> 16 # include <iostream> 17 # include <algorithm> 18 using namespace STD; 19/* Let's fight !!! */20 21 const int max_m = 524300; 22 typedef long ll; 23 ll ind [max_m], d [max_m], vis [max_m]; 24 ll T, H, n, m, a, B, IND; 25 # define rep (I, n) for (INT I = (0); I <(n); I ++) 26 void seed (LL & IND) {27 ind = (A * ind + B) % m; 28} 29 30 int main (void) {31 IOs :: sync_with_stdio (false); 32 CIN> T; 33 while (t --) {34 CIN> H> N> m> A> B> ind; 35 rep (I, m) CIN> d [I]; 36 37 int loop = 1; 38 mem Set (VIS,-1, sizeof (VIS); 39 vis [ind] = 0; 40 ind [0] = ind; 41 LL ans = 0; 42 while (loop <n) {43 ans + = d [ind]; 44 seed (IND); 45 if (vis [ind]! =-1) break; 46 vis [ind] = loop; 47 ind [loop] = ind; 48 + + loop; 49} 50 51 if (loop <= n-1) {52 LL sum = 0; 53 for (INT I = vis [ind]; I <loop; I ++) sum + = d [ind [I]; 54 LL times = (n-1-loop)/(loop-vis [ind]); 55 ans + = times * sum; 56 loop + = times * (loop-vis [ind]); 57 loop ++; 58} 59 While (loop <= n-1) {60 ans + = d [ind]; 61 seed (IND); 62 loop ++; 63} 64 65 cout <ans * H/2; 66 If (ANS * H % 2) cout <". 5 "; 67 else cout <". 0 "; 68 cout <Endl; 69} 70 71 return 0; 72}

Question C

To be continued...

Question d

To be continued...

Codechef August lunchtime 2014