whust #0.1 I

I - Laughing Out Loud

Time Limit:1000MS     Memory Limit:524288KB     64bit IO Format:%I64d & %I64u

Submit Status Practice Gym 100589I

Description

standard input/output 

Little Toojee was a happy go lucky boy. He seemed to find most, if not all, things funny. One day he read a word and started laughing a lot. Turns out that the word consisted only of the letters L and O. Whenever he saw the subsequence ‘LOL’ in the word, he laughed for 1 second. Given t strings, find out for how long Toojee laughed on seeing each string.

Input

The first line contains t queries. This is followed by t lines each containing one string S. String S consists only of capital alphabets.

Output

Output for each string on a new line.

Constraints

• 1  ≤ t ≤ 10
• 1  ≤ |S| ≤ 10⁵

Sample Input

Input

2
LOL
LOLOL

Output

1
4

Hint

Test1: On observation, we can tell that there is only 1 occurrence of LOL.

Test2: Let the string be 0-indexed and let V = {a, b, c} denote the indices that make up the string “LOL”, where a is index of the 1st ‘L’, b is index of the ‘O’ and c is the index of the 2nd ‘L’. So, V can be {0, 1, 2}, {2, 3, 4}, {0, 1, 4} and {0, 3, 4}. We see that there are 4 occurrences of the string “LOL”.

又是求有多少种组合的题.

和上次cf那道求等比数列的非常像.

/*************************************************************************
    > File Name: code/whust/#0.1/II.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月09日 星期日 16时00分44秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
int main()
{
    int T;
    cin>>T;
    string st;
    while (T--)
    {
    cin>>st;
    int len = st.length();
    LL p=0,q=0;
    LL ans = 0 ;
    for ( int i = 0 ; i < len ; i ++)
    {
        if (st[i]=='L')
        {
        p++;
        ans = ans + q;
        }
        else
        {
        q = q+ p; //p为L的个数,q表示前面有"LO"组合的个数
        }
    }
    cout<<ans<<endl;
    }
  
    return 0;
}