A. Vicious Keyboard Codeforce

最新推荐文章于 2018-09-27 15:57:05 发布

原创最新推荐文章于 2018-09-27 15:57:05 发布 · 947 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #codeforces

ACM-水题专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种算法，用于计算通过更改一个字符最多能获得多少次特定字符串VK作为子串出现的方法。输入为只包含V和K的字符串，输出则是最大出现次数。

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.

数据不大，直接暴力

Ac code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <map>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
#define fi first
#define se second
#define eps 1

const int maxn=200000+500;



int main() {  
	char s[105]; int vis[105];
	while(~scanf("%s",s)){
		
		memset(vis,0,sizeof vis);
		int cnt=0;
		int len=strlen(s);
		for(int i=0;i<len-1;i++){
			if(s[i]=='V'&&s[i+1]=='K'){
				cnt++;
				vis[i]=vis[i+1]=1;
				i++;
			}
		}
		for(int i=0;i<len-1;i++){
			if((s[i]=='V'||s[i+1]=='K')&&vis[i]==0&&vis[i+1]==0){
				cnt++;
				break;
			}
		}
		printf("%d\n",cnt );
	}
	return 0;
}