ZJ c462

// Judge: https://zerojudge.tw/ShowProblem?problemid=c462

#include <bits/stdc++.h>
using namespace std;

int nums[100000] = {0};
int bound = 0;

// My idea is to turn the string to several numbers, such as
// 'ABabaA' => [2,3,1]
// 'IJjiHW' => [2,2,2]
// 'dwFwGe' => [2,1,1,1,1]
// 'ojojowP' => [6,1]
// Got the pattern? The numbers show the amount of consecutive letters in the same case (both in lower or upper)
//
// Now the problem would be:
//     find the maximum length in these nums satisfy the conditions:
//         (nums[l] >= k, nums[r] >= k, nums[mid] == k, l < mid < r)
//     Let us call the nums[l] 'head', nums[mid] 'body', and nums[r] 'tail'.
int main(){
    // the variable k we declare
    int k;
    cin >> k;
    // the input string
    string s;
    cin >> s;
    // the variable 'cnt' represent the current number of same-case-letters
    int cnt = 0;
    // Remember, string's indexes start with 0, not 1
    for(int i = 0; i < s.size(); ++i){
        // s[i] is the current letter, and s[i-1] is the previous letter
        if(i == 0){
            // the first one, of course cnt is 1
            cnt = 1;
        }else if('a' <= s[i] && s[i] <= 'z'){
            if('a' <= s[i-1] && s[i-1] <= 'z'){
                // both s[i] and s[i-1] are lowercase letters
                cnt += 1;
            }else if('A' <= s[i-1] && s[i-1] <= 'Z'){
                // s[i] and s[i-1] are in different case
                nums[bound++] = cnt;
                cnt = 1;
            }
        }else if('A' <= s[i] && s[i] <= 'Z'){
            if('a' <= s[i-1] && s[i-1] <= 'z'){
                // s[i] and s[i-1] are in different case
                nums[bound++] = cnt;
                cnt = 1;
            }else if('A' <= s[i-1] && s[i-1] <= 'Z'){
                // both s[i] and s[i-1] are uppercase letters
                cnt += 1;
            }
        }
        if(i == s.size()-1){
            // This step is important!!!
            // At last one, you may miss to add cnt to nums[]
            nums[bound++] = cnt;
        }
    }
    // Here we have to find consecutive numbers as many as possible
    // Remember, the head and the tail of the sequence could be more than k letters
    int tail = 0; // 1 represents the current num could be the tail but not for body
    int ans = 0; // the current best answer
    int cur = 0; // the current answer
    for(int i = 0; i < bound; ++i){
        if(nums[i] == k){ // body
            if(tail == 1){
                cur = 2;
            }else{
                cur += 1;
            }
            tail = 0;
        }else if(nums[i] > k){ // head or tail
            if(cur == 0){ // no body or tail, so let nums[i] be head would be better
                cur += 1;
            }else if(tail == 1){ // previous tail be head, nums[i] be tail
                cur = 2;
            }else{ //
                cur += 1;
            }
            tail = 1;
        }else{ // nums[i] cannot be part of answer
            cur = 0;
            tail = 0;
        }
        // maintain the best answer
        ans = max(ans, cur);
    }
    cout << ans * k << '\n';
    return 0;
}
// It may seems that I wrote a LONG code, however I wrote lots of comments too.
// You may delete all these comments and see how short the code is :-)