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Oil tanker Permutations

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Unique Atm in wonderland Answer in python 3 a=int(input()) b=[] for i in range(a): temp=int(input()) b.append(temp) c=int(input()) if(sum(b)=0): sum1=sum1-b[i] i=i-1 r=r-1 else: break if(sum1==c and r

#include <iostream> #include <vector> #include <algorithm> using namespace std; int nSk(int n, int k){ if (n < 2) return 1; if (k > n-2) return 0; vector<int> numbers; numbers.reserve(n - 2); for (int i = 0; i < n - 2; ++i){ numbers.push_back(i + 2); } std::vector<bool> v(numbers.size()); std::fill(v.end() - k, v.end(), true); int ret = 0; do { int tmpnum = 1; for (int j = 0; j < numbers.size(); ++j) { if (v[j]) { tmpnum *= numbers[j]; } } ret += tmpnum; } while (std::next_permutation(v.begin(), v.end())); return ret; } int main(){ int n, d; cin >> n; cin >> d; int sum = 0; for (int k = 0; k < d; ++k){ sum += nSk(n, k); } //sum *= 2; cout<< sum; } ✅1 to N Friends puzzle code

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