A. Vicious Keyboard

A. Vicious Keyboard
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Tonio has a keyboard with only two letters, “V” and “K”.

One day, he has typed out a string s with only these two letters. He really likes it when the string “VK” appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times “VK” can appear as a substring (i. e. a letter “K” right after a letter “V”) in the resulting string.

Input
The first line will contain a string s consisting only of uppercase English letters “V” and “K” with length not less than 1 and not greater than 100.

Output
Output a single integer, the maximum number of times “VK” can appear as a substring of the given string after changing at most one character.

Examples
input
VK
output
1
input
VV
output
1
input
V
output
0
input
VKKKKKKKKKVVVVVVVVVK
output
3
input
KVKV
output
1
Note
For the first case, we do not change any letters. “VK” appears once, which is the maximum number of times it could appear.

For the second case, we can change the second character from a “V” to a “K”. This will give us the string “VK”. This has one occurrence of the string “VK” as a substring.

For the fourth case, we can change the fourth character from a “K” to a “V”. This will give us the string “VKKVKKKKKKVVVVVVVVVK”. This has three occurrences of the string “VK” as a substring. We can check no other moves can give us strictly more occurrences.

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<iomanip>#include<string>using namespace std;#define INF 9999999int main(){    string str;    while (cin >> str)    {        int ans = 0;int flag = 0;        for (int i = 0;i < str.length() - 1;i++)        {            if (str.at(i) == 'V')            {                if (str.at(i + 1) == 'K')                {                    ans++;                }            }        }         str = str+"  ";        string temp = "ABCD";            str = temp + str;            //cout << str << endl;        for (int i = 0;i < str.length()-2;i++)        {            if (str.at(i) == 'V'&&str.at(i + 1) == 'V'&&str.at(i + 2) != 'K')            {                flag = 1;                break;            }            else if (str.at(i) != 'V'&&str.at(i + 1) == 'K'&&str.at(i + 2) == 'K')            {                flag = 1;                break;            }        }        cout << ans + flag << endl;    }    return 0;}