Zuhair and Strings-祖海和字符串 CodeForce#1105B

题目链接：Zuhair and Strings

题目原文

Given a string 𝑠 of length 𝑛 and integer 𝑘 (1≤𝑘≤𝑛). The string 𝑠 has a level 𝑥, if 𝑥 is largest non-negative integer, such that it's possible to find in 𝑠:

𝑥 non-intersecting (non-overlapping) substrings of length 𝑘,
all characters of these 𝑥 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 𝑖 and 𝑗 (1≤𝑖≤𝑗≤𝑛), denoted as 𝑠[𝑖…𝑗] = "𝑠𝑖𝑠𝑖+1…𝑠𝑗".

For example, if 𝑘=2, then:

the string "aabb" has level 1 (you can select substring "aa"),
the strings "zzzz" and "zzbzz" has level 2 (you can select two non-intersecting substrings "zz" in each of them),
the strings "abed" and "aca" have level 0 (you can't find at least one substring of the length 𝑘=2 containing the only distinct character).
Zuhair gave you the integer 𝑘 and the string 𝑠 of length 𝑛. You need to find 𝑥, the level of the string 𝑠.

题目大意

找出字符串s里最多的连续k次出现的字母的出现次数，重叠部分的不算。

比如说aaabbaa，k=2。

找a的话：aaabbaa或者aaabbaa能找到两个。

找b的话：aaabbaa只有一个。

所以取最大值，2。

思路

用一个二维数组flag[x][y]，x表示字母（x = (int)char - 97 )。y表示出现连续字母的个数的最大值。

比如aaabbaa中，flag[0][] = {0, 0, 3, 0, 0, 0, 2}, flag[1][] = {0, 0, 0, 0, 2, 0, 0}。

最终取Σ(flag[x][y]%2)最大值即可。

题解

#include <iostream>
#include <cstring>

using namespace std;

int len, k, ans, mmax;
char c1, c2;

int flag[30][200010];

int main(int argc, char const *argv[])
{
#ifdef debug
    freopen("test.txt", "r", stdin);
#endif
    cin >> len >> k;
    for(int i = 0; i < len; i++)
    {
        cin >> c2;
        int index = c2 - 97;
        if(c2 == c1)
        {
            flag[index][i] = flag[index][i-1]+1;
            flag[index][i-1] = 0;
        }
        else
        {
            flag[index][i] = 1;
        }
        c1 = c2;
    }
    for (int i = 0; i < 30; ++i)
    {
        long long tmp = 0;
        for (int j = 0; j < len; ++j)
        {
            tmp += flag[i][j] / k;
        }
        if(tmp > mmax)
        {
            mmax = tmp;
        }
    }
    cout << mmax;
    return 0;
}