목록Project Euler (37)
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1 2 3 4 5 s=set([]) for i in range(2, 101) : for j in range(2, 101) : s.add(i**j) print(len(s)) cs C로 풀려면 (a^b)^c = a^(b^c)임을 고려하면 풀릴거같다..! 근데 귀찮..
12345678910111213int main() { IOS; llong sum = 1; llong last = 1; for (int i = 1; i
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 vector alist = { 2, 7, 61 }; llong mul(llong a, llong b, llong mod) { return (a * b) % mod; } llong bin_pow(llong a, llong p, llong mod) { if (p == 0) return 1 % mod; if (p & 1) return mul(a, bin_pow(a, p - 1, mod), mod); return bin_pow(mul(a, a, mod), p..
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 int solve(int n) { set p; p.insert(1); int num = 1; int numnum = -1; int cnt = 0; bool startCycle = false; while (1) { num *= 10; if (num % n == 0) return 0; if (num % n == numnum) return cnt; if (p.count(num % n) && startCycle == false) { startCycle = true; numnum = num % n; } num = num % n; if (startCycle) c..
1 2 3 4 5 6 7 8 9 prev = 1 prevprev= 1 for i in range (3,5000): cur = prev + prevprev if len(str(cur)) >= 1000: print(i) break prevprev = prev prev = cur cs 계속 파이썬 쓰니깐 양심이 찔리잖어,,
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 int main() { VI fact; fact.push_back(1); llong mul = 1; for (int i = 1; i fact[curFact]) { NUM -= fact[curFact]; cnt++; } else { break; } } rep(i, 0, 10) { if (vis[i] == 0 && cnt == 0) { cout
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 int main() { VI c; for (int i = 1; i