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#include <stdio.h> int fact(int num){ if(num>1){ return num*fact(num-1); } else if(num<=1){ return 1; } } int main() { int n,p,res; scanf("%d",&n); scanf("%d",&p); res = fact(n)/(fact(p)*fact(n-p)); printf("%d",res); return 0; } Pollution code https: http://t.me/Coding_solution_0

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def fib(n): if n <= 1: return n if n > 0: return fib(n-1) + fib(n-2) x=int(input()) y=int(input()) z=x+y print(fib(z)) http://t.me/Coding_solution_0

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def matrix(arr, m, ans): for i in range(m): for j in range(len(arr[i])): ele = arr[i][j] key = ele dic.setdefault(key, []) dic[key].append([i, j]) for k, v in dic.items(): count = 1 v.sort() for i in range(len(v) - 1): if v[i][0] == v[i + 1][0]: continue else: count += 1 if count == m: for j in v: r = j[0] c = j[1] ans[r][c] = 0 elif count < m / 2: for j in v: r1 = j[0] c1 = j[1] ans[r1][c1] = 1 else: for t in v: r2 = t[0] c2 = t[1] if r2 % 2 == 0: ans[r2][c2] = 1 else: ans[r2][c2] = 0 for i in range(len(ans)): zeros = 0 ones = 0 for j in range(len(ans[0])): if ans[i][j] == 0: zeros += 1 else: ones += 1 if zeros == ones: for k in range(len(ans[0])): if ans[i][k] == 0: ans[i][k] = 1 else: ans[i][k] = 0 elif zeros == 0 or ones == 0: continue else: if ones % 2 == 1: for j in range(len(ans[0])): r = 0 if ones != 0: ans[i][j] = 1 ones -= 1 r = 1 if ones > 0: t1 = m - j - 1 ans[i][t1] = 1 ones -= 1 else: if r == 1: continue if j <= t1: ans[i][j] = 0 else: break m = int(input()) arr = [] dup = [] for _ in range(m): dup = [t for t in input().split()] arr.append(dup.copy()) ans = arr.copy() dic = {} matrix(arr, m, ans) for i in range(len(ans)): for j in range(len(ans[0]) - 1): print(ans[i][j], end=" ") print(ans[i][-1]) Python MATRIX Code ✅Telegram- http://t.me/Coding_solution_0

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