Problem:


Score : 100 points

Problem Statement

Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins.
Find the probability of having more heads than tails.

Constraints

N is an odd number.

1 \leq N \leq 2999

p_i is a real number and has two decimal places.

0 < p_i < 1

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 \ldots p_N

Output

Print the probability of having more heads than tails.
The output is considered correct when the absolute error is not greater than 10^{-9}.

Sample Input 1

3
0.30 0.60 0.80

Sample Output 1

0.612

The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

Thus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096 + 0.036 = 0.612.

Sample Input 2

1
0.50

Sample Output 2

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

Sample Input 3

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3

0.3821815872



Sample input:
3
0.30 0.60 0.80


Sample output:
0.612

Submission status: Accepted

Solution code:
def coins(probabilities):
    n = len(probabilities)
    res = [0] * (n + 1)
    res[0] = 1

    for i, p in enumerate(probabilities, start=1):
        q = 1 - p
        carry = 0
        for j in range(i + 1):
            res[j], carry = res[j] * q + carry, res[j] * p

    return sum(res[(n + 1) // 2:])


def main():
    input()  # n
    probabilities = [float(x) for x in input().split()]
    return coins(probabilities)


print(main())

Submission status: Accepted

Solution code:
from collections import Counter,defaultdict,deque
from heapq import heappop,heappush,heapify
import sys,bisect,math,itertools,fractions,pprint
sys.setrecursionlimit(10**8)
mod = 10**9+7
mod2 = 998244353
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpln(n): return list(int(sys.stdin.readline()) for i in range(n))

n = inp()
a = list(map(float,input().split()))
dp = [[0] * (n+5) for i in range(n+5)]
dp[0][0] = 1
for i in range(n):
    for j in range(n+1):
        if j == 0:
            dp[i+1][j] = dp[i][j] * (1-a[i])
            continue
        dp[i+1][j] = dp[i][j-1] * a[i] + dp[i][j] * (1-a[i])
print(sum(dp[n][n//2+1:]))
# pprint.pprint(dp)

Submission status: Accepted

Solution code:
n = int(input())
p = [float(x) for x in input().split()]

dp = [[0 for _ in range(n+1)] for __ in range(n+1)]

dp[0][0] = 1

for i in range(1,n+1):
    for j in range(i+1):
        if j == 0:
            dp[i][j] = (1-p[i-1])*dp[i-1][j]
        else:
            dp[i][j] = (1-p[i-1])*dp[i-1][j] + p[i-1]*dp[i-1][j-1]

print(sum(dp[n][(n+1)//2:]))

Submission status: Accepted

Solution code:
def wolne_mnozenie_wielomianu(wielomian1, wielomian2):
    odp = [0.0] * (len(wielomian1) + len(wielomian2) - 1)
    for i in range(len(wielomian1)):
        for j in range(len(wielomian2)):
            odp[i + j] += wielomian1[i] * wielomian2[j]
    while len(odp) > 1 and odp[-1] == 0.0:
        odp.pop()
    return odp

def wyrownaj(wielomian1, wielomian2):
    if len(wielomian1) > len(wielomian2):
        wielomian2 = wielomian2 + [0.0] * (len(wielomian1) - len(wielomian2))
    if len(wielomian2) > len(wielomian1):
        wielomian1 = wielomian1 + [0.0] * (len(wielomian2) - len(wielomian1))
    return wielomian1, wielomian2

def dodawanie_wielomianu(wielomian1, wielomian2):
    wielomian1, wielomian2 = wyrownaj(wielomian1, wielomian2)
        
    odp = [0.0] * len(wielomian2)
    for i in range(len(wielomian1)):
        odp[i] = wielomian1[i] + wielomian2[i]
    while len(odp) > 1 and odp[-1] == 0.0:
        odp.pop()
    return odp

def odejmowanie_wielomianu(wielomian1, wielomian2):
    wielomian1, wielomian2 = wyrownaj(wielomian1, wielomian2)
        
    odp = [0.0] * len(wielomian2)
    for i in range(len(wielomian1)):
        odp[i] = wielomian1[i] - wielomian2[i]
    while len(odp) > 1 and odp[-1] == 0.0:
        odp.pop()
    return odp

def szybkie_mnozenie_wielomianu(wielomian1, wielomian2):
    wielomian1, wielomian2 = wyrownaj(wielomian1, wielomian2)
    if len(wielomian1) < 3:
        return wolne_mnozenie_wielomianu(wielomian1, wielomian2)
    k = len(wielomian1) // 2
    
    a = wielomian1[k:]
    b = wielomian1[:k]
    c = wielomian2[k:]
    d = wielomian2[:k]
    
    ac = szybkie_mnozenie_wielomianu(a, c)
    bd = szybkie_mnozenie_wielomianu(b, d)
    w = szybkie_mnozenie_wielomianu(dodawanie_wielomianu(a, b), dodawanie_wielomianu(c, d))
    w_ac_bd = odejmowanie_wielomianu(w, dodawanie_wielomianu(ac, bd))
    
    ac = [0.0] * 2 * k + ac
    w_ac_bd = [0.0] * k + w_ac_bd
    
    odp = dodawanie_wielomianu(ac, dodawanie_wielomianu(bd, w_ac_bd))
    while len(odp) > 1 and odp[-1] == 0.0:
        odp.pop()
    return odp
  

def wczytaj_liste(): 
    wczytana_lista = input()
    lista_znakow = wczytana_lista.split()
    ostateczna_lista = []
    for element in lista_znakow:
        ostateczna_lista.append(int(element))
    return ostateczna_lista

def wczytaj_rzeczywiste():
    wczytana_lista = input()
    lista_znakow = wczytana_lista.split()
    ostateczna_lista = []
    for element in lista_znakow:
        ostateczna_lista.append(float(element))
    return ostateczna_lista

def wymnoz_duzo_wielomianow(wielomiany):
    if len(wielomiany) == 1:
        return wielomiany[0]
    
    k = (len(wielomiany) + 1) // 2
    lewo = wymnoz_duzo_wielomianow(wielomiany[:k])
    prawo = wymnoz_duzo_wielomianow(wielomiany[k:])
    
    return szybkie_mnozenie_wielomianu(lewo, prawo)

def rzuc_grosza_wiedzminowi_natychmiast():
    N = wczytaj_liste()[0]
    prawdopodopienstwa = wczytaj_rzeczywiste()
    rzuty = []
    for moneta in prawdopodopienstwa: 
        rzut = [1 - moneta, moneta]
        rzuty.append(rzut)
    
    po_rzutach = wymnoz_duzo_wielomianow(rzuty)
    
    odp = sum(po_rzutach[N//2+1:])
    print(odp)
    
rzuc_grosza_wiedzminowi_natychmiast()

Submission status: Accepted

Solution code:
N, = map(int, input().split())
Ps = map(float, input().split())
dp = [0]*((N+1)**2)
dp[0] = 1
for i, p in enumerate(Ps):
    for j in range(N+1):
        dp[(i+1)*(N+1)+j] = p*dp[i*(N+1)+j-1] + (1-p)*dp[i*(N+1)+j]
r = 0
for i in range(N, N//2, -1):
    r += dp[N*(N+1)+i]
print(r)

Submission status: Accepted

Solution code:
N, = map(int, input().split())
Ps = map(float, input().split())
dp = [[0]*(N+1)for _ in range(N+1)]
dp[0][0] = 1
for i, p in enumerate(Ps):
    for j in range(N+1):
        dp[i+1][j] = p*dp[i][j-1] + (1-p)*dp[i][j]
r = 0
for i in range(N, N//2, -1):
    r += dp[N][i]
print(r)

Submission status: Accepted

Solution code:
def wczytaj_liste(): 
    wczytana_lista = input()
    lista_znakow = wczytana_lista.split()
    ostateczna_lista = []
    for element in lista_znakow:
        ostateczna_lista.append(int(element))
    return ostateczna_lista

def wczytaj_rzeczywiste():
    wczytana_lista = input()
    lista_znakow = wczytana_lista.split()
    ostateczna_lista = []
    for element in lista_znakow:
        ostateczna_lista.append(float(element))
    return ostateczna_lista

def rzuc_grosza_wiedzminowi_migusiem():
    N = wczytaj_liste()[0]
    prawdopodopienstwa = wczytaj_rzeczywiste()
    przed_rzutem = [1.0]
    for moneta in prawdopodopienstwa: 
        po_rzucie = []
        for i in range(len(przed_rzutem)):
            if i > 0: 
                po_rzucie.append(przed_rzutem[i]*(1-moneta)+przed_rzutem[i-1]*(moneta))
            else:
                po_rzucie.append(przed_rzutem[0]*(1-moneta))
        po_rzucie.append(przed_rzutem[-1]*moneta)
        przed_rzutem = po_rzucie
    odp = sum(przed_rzutem[N//2+1:])
    print(odp)
    
rzuc_grosza_wiedzminowi_migusiem()

Submission status: Accepted

Solution code:
# input
N = int(input())
Ps = list(map(float, input().split()))

# probs[i][k] : コインiまでを投げて表がk枚出る確率
probs = [[0 for _ in range(N+1)] for _ in range(N)]

probs[0][1] = Ps[0]
probs[0][0] = 1 - Ps[0]
# for i in range(2, N+1):
#   probs[0][i] = 0

for i in range(1, N):
  for k in range(0, N+1):
    if k == 0:
      probs[i][k] = probs[i-1][k]*(1-Ps[i])
    else:
      probs[i][k] = probs[i-1][k-1]*Ps[i] + probs[i-1][k]*(1-Ps[i])

# print(probs)

sum_all = sum(probs[N-1])
sum_ok = 0

start = int(N/2) + 1
for i in range(start, N+1):
  sum_ok += probs[N-1][i]

print(sum_ok / sum_all)

Submission status: Accepted

Solution code:
n = int(input())
p = list(map(float, input().split()))

dp = [[0] * (n+1) for _ in range(n+1)]
dp[0][0] = 1

for i in range(1,n+1):
    dp[i][0] = dp[i-1][0] * (1- p[i-1])
    for j in range(1,n+1):
        dp[i][j] = (dp[i-1][j] * (1- p[i-1]) +
                    dp[i-1][j-1] * p[i-1])

ans = 0
for i in range((n//2 + 1) , n+1):
    ans += dp[n][i]

print(ans)

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    P = [float(x) for x in input().split()]
    over = (N + 1) // 2
    dp = [[0.0] * (over + 1) for _ in range(N + 1)]
    dp[0][0] = 1.0
    for i in range(N):
        for j in range(over):
            dp[i+1][j] += dp[i][j] * (1 - P[i])
            dp[i+1][j+1] += dp[i][j] * P[i]
        dp[i+1][over] += dp[i][over]
    print("{:.20f}".format(dp[-1][-1]))


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
n = int(input())
p = list(map(float, input().split()))

dp = [[0]*(n+1) for i in range(n+1)]
dp[0][0] = 1
for i in range(n):
    for j in range(n+1):
        if j-1 < 0:
            dp[i+1][j] = (1 - p[i]) * dp[i][j]
        else:
            dp[i+1][j] = p[i] * dp[i][j-1] + (1 - p[i]) * dp[i][j]

print(sum(dp[n][(n+1)//2:]))

Submission status: Accepted

Solution code:
from math import ceil

N = int(input())
p = [float(i) for i in input().split()]

# dp[i][j]: i枚を投げて、表がj枚だったときの確率
dp = [[0] * (N+1) for _ in range(N+1)]

# 初期条件
dp[0][0] = 1.0

# 配るDP
for i in range(N):
    for j in range(i+1):
        # 次のコイン(i+1)が表
        dp[i+1][j+1] += dp[i][j] * p[i]
        # 次のコイン(i+1)が裏
        dp[i+1][j] += dp[i][j] * (1-p[i])

answer = 0.0
for i in range(ceil(N/2),N+1):
    answer += dp[N][i]

print(answer)

Submission status: Accepted

Solution code:
n = int(input())
p = list(map(float, input().split()))
q=[]
q.append([1])
for i in range(n):
    q[0].append(0)
prod = 1.0
for i in range(1,n+1):
    prod*=p[i-1]
    q.append([prod])

for i in range(1,n+1):
    for j in range(1,n+1):
        try:
            q[i].append(p[i-1]*q[i-1][j] + (1-p[i-1])*q[i-1][j-1])
        except:
            print(i)
            print(j)
        #print(q)
ans = 0.0
for i in range(0,(n-1)//2+1):
    ans+=q[n][i]
print(ans)

Submission status: Accepted

Solution code:
N = int(input())
p = list(map(float, input().split()))

dp = []
for i in range(N+1):
	z = []
	for j in range(N):
		z.append(0)
	dp.append(z)

dp[0][0] = 1-p[0]

for i in range(N-1):
	dp[0][i+1] = dp[0][i]*(1-p[i+1])

dp[1][0] = p[0]

for i in range(N+1):
	for j in range(N-1):
		if j >= i-1 :
			dp[i+1][j+1]=p[j+1]*dp[i][j] + (1-p[j+1])*dp[i+1][j]
k = (N//2)+1
sump = 0
while(k<=N):
	sump += dp[k][N-1]
	k+=1
print(sump)

Submission status: Accepted

Solution code:
n = int(input())
p = [float(x) for x in input().split()]
dp = []
for i in range(n+1):
    tp = []
    for j in range(n+1):
        tp.append(0)
    dp.append(tp)
dp[0][1] =  p[0]
dp[0][0] = 1-p[0]
for i  in range(n):
    if i<=0:
        continue
    dp[i][0] = dp[i-1][0]*(1-p[i])
    for j in range(i+1):
        dp[i][j+1] = dp[i-1][j]*p[i]
        dp[i][j+1] += dp[i-1][j+1]*(1-p[i])

k = int(n/2)+1
ans = 0
#print(dp)
for j in range(k, n+1):
    ans += dp[n-1][j]
print(ans)

Submission status: Accepted

Solution code:
n = int(input())
probs = [float(c) for c in input().split()]

dp = [0] * (n+1)
dp[0] = 1
for coin in range(n):
    for heads in reversed(range(n+1)):
        dp[heads] = dp[heads] * (1 - probs[coin-1])
        if heads > 0:
            dp[heads] += dp[heads-1] * probs[coin-1]

ans = 0
for heads in range(n+1):
    tails = n - heads
    if heads > tails:
        ans += dp[heads]

print(ans)

Submission status: Accepted

Solution code:
n = int(input())
probs = [float(c) for c in input().split()]

dp = [[0] * (n+1) for _ in range(n+1)]
dp[0][0] = 1
for coin in range(1, n+1):
    for heads in range(n+1):
        dp[coin][heads] = dp[coin-1][heads] * (1 - probs[coin-1])
        if heads > 0:
            dp[coin][heads] += dp[coin-1][heads-1] * probs[coin-1]

ans = 0
for i in range(n//2 + 1, n+1):
    ans += dp[n][i]

print(ans)

Submission status: Accepted

Solution code:
N = int(input())
P = list(map(float,input().split()))
P.insert(0,0)
dp = [[0 for _ in range(N+1)] for _ in range(N+1)]
dp[1][0] = 1-P[1]
dp[1][1] = P[1]
for i in range(2,N+1):
    for j in range(N+1):
        dp[i][j] += dp[i-1][j]*(1-P[i])
        if j>=1:
            dp[i][j] += dp[i-1][j-1]*P[i]
cnt = 0
for i in range(N//2+1,N+1):
    cnt += dp[N][i]
print(cnt)

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    N = r_int()
    P = Rfloat()
    dp = [[0] * (N+1) for _ in range(N+1)]
    dp[0][0] = 1
    for j, p in enumerate(P):
        for i in range(j+1):
            dp[i][j+1] += dp[i][j] * (1-p)
            dp[i+1][j+1] += dp[i][j] * (p)
    print(sum(dp[i][N] for i in range(N//2+1, N+1)))
            
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
from math import ceil
input = lambda: sys.stdin.readline().rstrip()
tdlist = lambda i, j, value=0: [[value] * j for _ in range(i)]

n = int(input())
p = list(map(float, input().split()))
dp = tdlist(n+1, n+1)
dp[0][0] = 1

for i in range(1, n+1):
    for j in range(n+1):
        if j > 0: dp[i][j] += dp[i-1][j-1] * p[i-1]
        dp[i][j] += dp[i-1][j] * (1 - p[i-1])

print(sum(dp[n][ceil(n/2):]))

Submission status: Accepted

Solution code:
n = int(input())
p = list(map(float, input().split()))
dp = [0] * (n + 1) # dp[i]: i tosses, probability that there are i heads so far
dp[0] = 1
for coin in range(n):
    for i in range(coin + 1, -1, -1):
        if i - 1 >= 0:
            dp[i] = dp[i - 1] * p[coin] + dp[i] * (1 - p[coin])
        else:
            dp[i] = dp[i] * (1 - p[coin])
res = 0
for heads in range(n + 1):
    tails = n - heads
    if heads > tails:
        res += dp[heads]
print(res)

Submission status: Accepted

Solution code:
n = int(input())
arr = list(map(float, input().split()))
dp = [0] * (n + 1)
# Base Case
dp[0] = 1

# 0/1 knapsack sarika, pick karna hai ya na karna hai vaisa hi, head aayega ya nahi, aayega to uski prob, + nahi aayega to uski prob, dono ka combi wrt to the last combination we already dealt with.

for i, j in enumerate(arr):
    for k in range(i + 1, -1, -1):
        # if k == 0, i.e. no heads, so head ki prob nahi dalni isme, otherwise combination aayega head aur tails ki prob ka
        if k:
            dp[k] = (dp[k - 1] * j) + (dp[k] * (1-j))
        else:
            dp[k] = dp[k] * (1-j)

mid = n // 2 + 1
ans = 0
for i in range(mid, len(arr) + 1):
    ans += dp[i]
print("{:.10f}".format(ans))

Submission status: Accepted

Solution code:
N = int(input())
P = list(map(float, input().split()))
dp = [[0] * (N+1) for _ in range(N+1)]
dp[0][0] = 1
for i in range(N):
    p = P[i]
    for j in range(i+1):
        dp[i+1][j+1] += dp[i][j] * p
        dp[i+1][j] += dp[i][j] * (1-p)

ans = sum(dp[N][i] for i in range(N//2+1, N+1))
print("{:.10f}".format(ans))

Submission status: Accepted

Solution code:
N = int(input())
P = [float(i) for i in input().split()] 
 
dp = [[0 for i in range(N+1)] for j in range(N+1)]
dp[0][0] = 1
 
for i in range(1,N+1):
    for j in range(0,i+1):
        if j == 0:
            dp[i][j] = dp[i-1][j] * (1-P[i-1])
        elif j == i:
            dp[i][j] = dp[i-1][j-1] * P[i-1]
        else:
            dp[i][j] = dp[i-1][j]*(1-P[i-1]) + dp[i-1][j-1]*P[i-1]
print(sum(dp[-1][N//2 + 1:]))