Problem:
Score : 300 points

Problem Statement

There are N Reversi pieces arranged in a row. (A Reversi piece is a disc with a black side and a white side.)
The state of each piece is represented by a string S of length N.
If S_i=B, the i-th piece from the left is showing black;
If S_i=W, the i-th piece from the left is showing white.

Consider performing the following operation:

Choose i (1 \leq i < N) such that the i-th piece from the left is showing black and the (i+1)-th piece from the left is showing white, then flip both of those pieces. That is, the i-th piece from the left is now showing white and the (i+1)-th piece from the left is now showing black.

Find the maximum possible number of times this operation can be performed.

Constraints

1 \leq |S| \leq 2\times 10^5

S_i=B or W

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum possible number of times the operation can be performed.

Sample Input 1

BBW

Sample Output 1

2

The operation can be performed twice, as follows:

Flip the second and third pieces from the left.

Flip the first and second pieces from the left.

Sample Input 2

BWBWBW

Sample Output 2

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int n,arr[200001];
int ans=1987654321;

vector <pair<int,int> > before;

bool f(int x)
{
    before.clear();
    if(x<=1) return false;
    for(int i=1;i<n;i++)
    {
        if(arr[i-1]<arr[i]) continue;
        while(!before.empty() && before.back().first>arr[i]) before.pop_back();
        for(int j = arr[i]-1 ; ; j--)
        {
            if(j<0) return false;
            if(before.empty())
            {
                before.push_back({j,1});
                break;
            }
            pair <int,int> t = before.back();
            before.pop_back();
            if(t.first == j)
            {
                if(t.second>=x) continue;
                t.second++;
                before.push_back(t);
            }
            else
            {
                before.push_back(t);
                before.push_back({j,1});
            }
            break;
        }
    }
    return true;
}

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d",&arr[i]);
    int s=1,e=n;
    for(s=1,e=n;s<=e;)
    {
        int m=(s+e)/2;
        if(f(m)) ans=m,e=m-1;
        else s=m+1;
    }
    printf("%d",ans);
}