My solution for the Problem A of the 2017's edition of Google Code Jam: Oversized Pancake Flipper

gcj-2017-A.py

#! /usr/bin/env python

import sys


def solve(line):
    pancake_row = [p == '+' for p in line.split()[0]]
    pan_size = int(line.split()[1])

    flips = 0
    i = 0
    while i < (len(pancake_row) - pan_size + 1):
        if not pancake_row[i]:
            for j in range(i,i + pan_size):
                pancake_row[j] = not pancake_row[j]
            flips += 1
        i += 1

    if all(pancake_row):
        return flips
    return 'IMPOSSIBLE'


def progress(s):
    print("%-80s\r" % s, file=sys.stderr, end='')


if __name__ == '__main__':
    if len(sys.argv) > 1:
        inputfile = sys.argv[1]
    else:
        inputfile = 'input.test'
    with open(inputfile) as f:
        cases = int(f.readline())
        for i in range(cases):
            progress(i)
            line = f.readline().strip()
            print('Case #%d: %s' % (i+1, solve(line)))

You don't totally understand the question well. This is a state transfer machine problem. Naive solution consists of Wide First Search(wfs) and mathematical branches trimming. You loop though the array from left to right? I don't believe this is a accept answer.

here is simple mathematics about trmming:
K * flips = x* 2p + y * (2q-1). Where x represents the number of '+' in initial state and y represents the number of '-'. A Filter can be summarised as

int
math_rule(const char* arr, int l, int K){
    int i, x=0, y=0;
    for (i=0; i < l; i++) {
        if (arr[i] == '+') {
            x += 1;
        } else
        if (arr[i] == '-') {
            y += 1;
        }
    }
    
    // K*n = x*2*k1 + y*(2*k2 - 1)
    if (x == y and K % 2 == 0 and x % K != 0) {
        return IMPOSSIBLE;
    }
    
    return 0;
}

each state can represented by a bit representation or an unique integer. I am serious. Google's problems need to you to think twice before coding.

Consider the case ++-+-++ 3
i don't see any possible solution for this.
here x=5,y=2,k=3.
here x!=y, k%2!=0
So by your logic, its not impossible.
Could you please explain how to flip all values?