hihocoder#1052 Genetic Engineering

Original title Address

Find a regular problem

If it is not, then the number of different digits is the number of times the change is required.

If the end-to-end intersection seems more complex, so look for patterns.

Suppose the string is this (the same string up and down, just to make it easier to describe the end-to-end part), so that the blue end-to-end part is the same:

Because it is a string, the dashed parts are the same.

That is, the middle Red box section is the same.

Because the blue and the end parts are the same, there are:

The dashed part is the same:

And at this point, we can get the same dashed part:

Finally, repeating this process, we divided the blue part into a follow-up link:

As you can see from the first step, the length of this cycle section is N-k

So, as long as the blue part is n-k for the length of the loop can satisfy the condition.

Code:

1#include <iostream>2#include <cstring>3 4 using namespacestd;5 6 intMain () {7   intT, K, N;8   Charstr[1024x768];9 TenCIN >>T; One    while(t--) { ACIN >>str; -CIN >>K; -N =strlen (str); the     intres =0; -  -     ifK2<=N) { -        for(inti =0, j = n-k; J < N; J + +, i++) +Res + = Str[i]! = str[j]?1:0; -cout << Res <<Endl; +     } A     Else { at       intLen = N-K; -        for(inti =0; i < Len; i++) { -         intcount[4] = {0}; -          for(intj = i; J < N; J + =Len) { -           if(Str[j] = ='A') -count[0]++; in           Else if(Str[j] = ='T') -count[1]++; to           Else if(Str[j] = ='C') +count[2]++; -           Else if(Str[j] = ='G') thecount[3]++; *         } $         intMaximum =0;Panax Notoginseng          for(intj =0; J <4; J + +) { -Maximum =max (maximum, count[j]); theRes + =Count[j]; +         } ARes-=maximum; the       } +cout << Res <<Endl; -     } $   } $  -   return 0; -}

hihocoder#1052 Genetic Engineering