Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 6005 Accepted: 2777

Description

Farmer John has arranged his N (1 ≤ N ≤ 5,000) cows in a row and many of them are facing forward, like good cows. Some of them are facing backward, though, and he needs them all to face forward to make his life perfect.

Fortunately, FJ recently bought an automatic cow turning machine. Since he purchased the discount model, it must be irrevocably preset to turn K (1 ≤ K  N) cows at once, and it can only turn cows that are all standing next to each other in line. Each time the machine is used, it reverses the facing direction of a contiguous group of K cows in the line (one cannot use it on fewer than K cows, e.g., at the either end of the line of cows). Each cow remains in the same *location* as before, but ends up facing the *opposite direction*. A cow that starts out facing forward will be turned backward by the machine and vice-versa.

Because FJ must pick a single, never-changing value of K, please help him determine the minimum value of K that minimizes the number of operations required by the machine to make all the cows face forward. Also determine M, the minimum number of machine operations required to get all the cows facing forward using that value of K.

Input

Line 1: A single integer: N 
Lines 2..N+1: Line i+1 contains a single character, F or B, indicating whether cow i is facing forward or backward.

Output

Line 1: Two space-separated integers: K and M

Sample Input

7
B
B
F
B
F
B
B

Sample Output

3 3

Hint

For K = 3, the machine must be operated three times: turn cows (1,2,3), (3,4,5), and finally (5,6,7)
呐 题意就这样
然后我们考虑枚举k做一个O(n^3)显然很简单
优化?尺取法
枚举k,从左到右对背面的牛操作
考虑能够影响当前牛的操作,是当前减去(k-1)的区间
然后统计一下这个区间有多少个操作对2取模就可以得到当前判断。
这样用一个类似前缀和的东西维护就行了 见代码
O(n^2)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<iterator>
#include<cstdlib>
using namespace std;

bool cow[10010]={0},cowx[10010]={0},opt[10010]={0};
int n,m,ans=2*1e9,ansx,sum[10010]={0};
char ch1=0; 

inline void readx(int& x)
{
	x=0; int k=1; register char ch=0;
	while (ch<'0' || ch>'9') { ch=getchar(); if (ch=='-') k=-1; }
	while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	x*=k;
}

inline bool judge()
{
	for (int i=n+1;i<=2*n;i++) if (!cow[i]) return false;
	return true; 
}

int main()
{
	readx(n);
	for (int i=n+1;i<=2*n;i++)
	{
		ch1=getchar();
		while (ch1!='F' && ch1!='B') ch1=getchar();
		if (ch1=='F') cowx[i]=true;
		else cowx[i]=false;
	}
	
	for (int k=1;k<=n;k++)
	{
		memset(opt,0,sizeof(opt));
		memset(sum,0,sizeof(sum));
		memcpy(cow,cowx,sizeof(cowx));
		m=0;
		for (int i=n+1;i<=2*n;i++)
		{
			sum[i]=sum[i-1]+opt[i-1]-opt[i-k];
			if (sum[i]%2) cow[i]=cow[i]^1;
			
			if (i<=2*n-k+1 && (!cow[i]))
			{
				cow[i]=true;
				opt[i]=true;
				m++;
			}
		}
		if (judge() && ans>m)
		{
			ans=m; ansx=k;
		}
	}
	printf("%d %d\n",ansx,ans);
	return 0;
}

发表评论

电子邮件地址不会被公开。 必填项已用*标注

This site uses Akismet to reduce spam. Learn how your comment data is processed.