Codeforces 353D Queue【思维】好题~

最新推荐文章于 2024-05-29 10:30:01 发布

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最新推荐文章于 2024-05-29 10:30:01 发布

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分类专栏：思维文章标签： Codeforces 353D

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本文探讨了一个关于队列的问题，即如何通过交换位置的方式将所有女生(F)置于男生(M)之前，并提供了一种有效的算法来计算所需的最短时间。

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There are n schoolchildren, boys and girls, lined up in the school canteen in front of the bun stall. The buns aren't ready yet and the line is undergoing some changes.

Each second all boys that stand right in front of girls, simultaneously swap places with the girls (so that the girls could go closer to the beginning of the line). In other words, if at some time the i-th position has a boy and the (i + 1)-th position has a girl, then in a second, the i-th position will have a girl and the (i + 1)-th one will have a boy.

Let's take an example of a line of four people: a boy, a boy, a girl, a girl (from the beginning to the end of the line). Next second the line will look like that: a boy, a girl, a boy, a girl. Next second it will be a girl, a boy, a girl, a boy. Next second it will be a girl, a girl, a boy, a boy. The line won't change any more.

Your task is: given the arrangement of the children in the line to determine the time needed to move all girls in front of boys (in the example above it takes 3 seconds). Baking buns takes a lot of time, so no one leaves the line until the line stops changing.

Input

The first line contains a sequence of letters without spaces s₁s₂... s_n (1 ≤ n ≤ 10⁶), consisting of capital English letters M and F. If letter s_i equals M, that means that initially, the line had a boy on the i-th position. If letter s_i equals F, then initially the line had a girl on the i-th position.

Output

Print a single integer — the number of seconds needed to move all the girls in the line in front of the boys. If the line has only boys or only girls, print 0.

Examples

Input

MFM

Output

Input

MMFF

Output

Input

FFMMM

Output

Note

In the first test case the sequence of changes looks as follows: MFM → FMM.

The second test sample corresponds to the sample from the statement. The sequence of changes is: MMFF → MFMF → FMFM → FFMM.

题目大意：

F表示女生，M表示男生，每秒如果一个女生前边有一个男生，两个人就可以进行调换位子，问一共需要多少秒，所有的F都在左边，所有的M都在右边。

思路（思路源自：http://blog.youkuaiyun.com/w446506278/article/details/52068170?locationNum=2）：

对于一个女生，其肯定要交换的次数为她前边的男生的个数次。

那么如果多个女生连在了一起，其会有延时的可能，那么对应我们统计这个女生前边的男生的个数。初始耗时为0.遇到的女生，如果其前边男生的个数不为0.那么这个女生走到应该到的位子需要的耗时为max（他前边男生的个数，ans+1）；

也就是说，如果前边男生的个数比较少的时候，会有延时的情况发生，如果他前边的男生的个数足够多，那么其不会发生延时的情况。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char a[1000060];
int main()
{
    while(~scanf("%s",a))
    {
        int n=strlen(a);
        int ans=0;
        int contm=0;
        for(int i=0;i<n;i++)
        {
            if(a[i]=='M')contm++;
            else
            {
                if(contm>0)
                {
                    ans=max(ans+1,contm);
                }
            }
        }
        printf("%d\n",ans);
    }
}