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傳送門：http://codeforces.com/contest/1105/problem/B

B. Zuhair and Strings time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Given a string ss of length nn and integer kk (1≤k≤n1≤k≤n). The string ss has a level xx, if xx is largest non-negative integer, such that it‘s possible to find in

ss:

• xx non-intersecting (non-overlapping) substrings of length kk,
• all characters of these xx substrings are the same (i.e. each substring contains only one distinct character and this character is the same for all the substrings).

A substring is a sequence of consecutive (adjacent) characters, it is defined by two integers

ii and jj (1≤i≤j≤n1≤i≤j≤n), denoted as s[i…j]s[i…j]= "sisi+1…sjsisi+1…sj".

For example, if k=2k=2, then:

• the string "aabb" has level 11 (you can select substring "aa"),
• the strings "zzzz" and "zzbzz" has level 22 (you can select two non-intersecting substrings "zz" in each of them),
• the strings "abed" and "aca" have level
  00 (you can‘t find at least one substring of the length k=2k=2 containing the only distinct character).

Zuhair gave you the integer kk and the string ss of length nn. You need to find xx, the level of the string ss.

Input

The first line contains two integers nn and kk (1≤k≤n≤2?1051≤k≤n≤2?105) — the length of the string and the value of kk.

The second line contains the string ss of length nn consisting only of lowercase Latin letters.

Output

Print a single integer xx — the level of the string.

Examples input Copy

8 2
aaacaabb

output Copy

2

input Copy

2 1
ab

output Copy

1

input Copy

4 2
abab

output Copy

0

Note

In the first example, we can select 22 non-intersecting substrings consisting of letter ‘a‘: "(aa)ac(aa)bb", so the level is 22.

In the second example, we can select either substring "a" or "b" to get the answer 11.

題意概括：

給一串長度為 N 的字符，求長度為 K 的由相同單種字母組成的子串的最大數量。

解題思路：

單指針模擬長度為 K 的子串的起始點，O(N)遍歷一遍整個字符，記錄每種字母滿足條件的子串所對應的數量。

AC code：

 1 #include <bits/stdc++.h>
 2 #define INF 0x3f3f3f3f
 3 #define LL long long
 4 using namespace std;
 5 const int MAXN = 2e5+10;
 6 char str[MAXN];
 7 int num[30];
 8 int main()
 9 {
10     int N, K;
11     scanf("%d %d", &N, &K);
12     scanf("%s", str);
13     int len = strlen(str);
14     int st = 0;
15     bool flag;
16     while(st+K-1 < len){
17         flag = true;
18         for(int i = st+1; i <= st+K-1; i++){
19             //printf("%c %c\n", str[i], str[i-1]);
20             if(str[i] != str[i-1]){
21                 flag = false;
22                 st = i;
23                 break;
24             }
25         }
26         if(flag){
27             //printf("%d %c\n",st, str[st]);
28             num[str[st]-‘a‘]++;
29             st = st+K;
30         }
31         //st+=K;
32     }
33     int ans = 0;
34     for(int i = 0; i < 26; i++){
35         //printf("%d %d\n", i, num[i]);
36         ans = max(num[i], ans);
37     }
38 
39     printf("%d\n", ans);
40     return 0;
41 
42 }

Codeforces Round #533 (Div. 2) B. Zuhair and Strings 【模擬】