O‘zbekiston respublikasi raqamli texnologiyalar vazirligi muhammad




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Task 9. Farida


O‘ZBEKISTON RESPUBLIKASI RAQAMLI TEXNOLOGIYALAR VAZIRLIGI
MUHAMMAD AL-XORAZMIY NOMIDAGI TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI





Dasturiy injiniring yo‘nalish talabalasi Quronboyeva Faridaning

bajargan

AMALIY ISHI









TOSHKENT- 2023

Task 9. Zinapoya N zinadan iborat. Quyon bir sakrashda K dan ortiq qadam bosib o’ta olmaydi. K va N ning qiymatlari berilganda quyon zinapoyaning tepasiga necha xil yo’l bilan chiqishi mumkin.

Masalan, agar K=3 va N=4 bo’lsa, unda quyidagi yo’llar mavjud: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+ 3, 3+1. Demak, Output:
7 Input: K=2, N=7; Output: 21

def quyon_tepasiga_chiqish(K, N):


if K == 1:
return 1
if N == 1:
return 1
if N < K:
return 0

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


for i in range(K + 1):
dp[i][0] = 1

for i in range(1, K + 1):


for j in range(1, N + 1):
dp[i][j] = dp[i][j - 1] + dp[i - 1][j]

return dp[K][N]


K = int(input("K ni kiriting: "))


N = int(input("N ni kiriting: "))

result = quyon_tepasiga_chiqish(K, N)


print("Output:", result)

def count_ways_to_top(K, N):


# Create a list to store the number of ways to reach each step
ways = [0] * (N + 1)
# Initialize the base cases
ways[0] = 1
ways[1] = 1
# Compute the number of ways for each step up to N
for i in range(2, N + 1):
for j in range(1, K + 1):
if i - j >= 0:
ways[i] += ways[i - j]
# Return the number of ways to reach the top
return ways[N]

# Test the function


K = 3
N = 4
print(count_ways_to_top(K, N)) # Output: 7

K = 2
N = 7


print(count_ways_to_top(K, N)) # Output: 21
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O‘zbekiston respublikasi raqamli texnologiyalar vazirligi muhammad

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