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  1. #include <bits/stdc++.h>
  2. #include<unordered_set>
  3. #define el "\n"
  4. #define ll long long
  5. using namespace std;
  6. #define yes cout<<"YES\n";
  7. #define no cout<<"NO\n";
  8. #define all(v) v.begin(), v.end()
  9. #define rall(a) a.rbegin() , a.rend() 
  10. #define sz(a)  a.size()
  11. const ll N = 1e7 + 5, MOD = 1e9 + 7;
  12. void ELMOO() {
  13.     ios_base::sync_with_stdio(false);
  14.     cin.tie(nullptr);
  15. }
  16. bool isPalindrome(ll x) {                 // function Palindrome 
  17.     string s = to_string(x);
  18.     string r = s;
  19.     reverse(all(r));
  20.     return s == r;
  21. }
  22. //-------------------------------------------------------------------------------------------
  23. // o(sqrt(n)) ---> max_n = 10^12 ~ 10^15    --- Function to find the number is prime or not
  24. bool is_Prime(ll n) {
  25. 	if (n < 2)return 0;
  26.     for (ll i = 2; i <= n/i; i++)
  27.     {
  28.         if (n % i == 0)return 0;
  29.     }
  30.     return 1;
  31. }
  32. //-------------------------------------------------------------------------------------------
  33. // O(sqrt(n)) ---> max_n = 10^12 ~ 10^15    --- Function to find all divisors of a number
  34. vector<ll> factorize(ll n) {
  35.     vector<ll> divisors;
  36.     for (ll i = 1; i <= n / i; i++)
  37.     {
  38.         if(n%i==0) {
  39.             divisors.push_back(i);
  40.             if(i*i != n) {
  41.                 divisors.push_back(n/i);
  42.             }
  43. 		}
  44.     }
  45.     return divisors;
  46. }
  47. //---------------------------------------------------------------------------------------------
  48. // O(sqrt(n)) ---> max_n = 10^12 ~ 10^15    --- Function to find all prime factors of a number
  49. vector<ll>Prime_Factorize(ll n) {
  50.     vector<ll> primes;
  51.     for (ll i = 2; i <= n / i; i++)
  52.     {
  53.         while (n % i == 0) {
  54.             primes.push_back(i);
  55.             n /= i;
  56.         }
  57.     }
  58. 	if (n > 1)primes.push_back(n);// لو عدد اكبر من الجذر التربيعي يبقى هو نفسه عدد اولي
  59.     return primes;
  60. }
  61. //---------------------------------------------------------------------------------------------
  62. // O(n log log n)  --- max_n = 10^7    --- Function to find all prime numbers up to n
  63. vector<bool> is_prime(N, 1);
  64. void sieve() {
  65.     is_prime[0] = is_prime[1] = 0;
  66.     for (ll i = 2;i <= N / i;i++) {
  67.         if (is_prime[i] ) {
  68.             for (ll j = i * i;j <= N;j += i) {
  69.                 is_prime[j] = 0;
  70.             }
  71.         }
  72.     }
  73. }
  74. //---------------------------------------------------------------------------------------------
  75. // O(n log n)  --- max_n = 10^5    --- Function to find the number of divisors for each number up to n
  76. vector<ll>nm_of_dv(N);
  77. void sieve_of_divisors() {
  78.     for (ll i = 1;i <= N;i++) {
  79.         for (ll j = i;j < N;j += i) {
  80.             nm_of_dv[j]++;
  81.         }
  82.     }
  83. }
  84. //---------------------------------------------------------------------------------------------
  85. // O(n log n)  --- max_n = 10^7    --- Function to find the smallest prime factor for each number up to n
  86. vector<int>sp(N+1);
  87. void spf() {
  88.     for (int i = 0; i <= N; i++)sp[i] = i;
  89.     for (int i = 2;i <= N / i;i++) {
  90.         if (sp[i] == i) {
  91.             for (int j = i * i;j <= N;j += i) {
  92. 				if (sp[j] == j)sp[j] = i;
  93. 				//sp[j] = min(sp[j], i);
  94.             }
  95.         }
  96.     }
  97. }
  98. //---------------------------------------------------------------------------------------------
  99. // O(log n)  --- max_n = 10^12 ~ 10^15    --- Function to find the GCD of two numbers
  100. ll GCD(ll a, ll b) {
  101. 	if (b > a)swap(a, b);
  102. 	if (!b)return a;
  103. 	return GCD(b, a % b);
  104. }
  105. //---------------------------------------------------------------------------------------------
  106. // O(log n)  --- max_n = 10^12 ~ 10^15    --- Function to find the LCM of two numbers
  107. ll LCM(ll a, ll b) {
  108.     return (a / GCD(a, b)) * b;
  109. }
  110. //---------------------------------------------------------------------------------------------
  111. // O(log n)  --- max_n = 10^12 ~ 10^15    --- Function to find the sum of two numbers under modulo
  112. ll Mod_add(ll a, ll b) {
  113.     return ((a % MOD) + (b % MOD)) % MOD;
  114. }
  115. //---------------------------------------------------------------------------------------------
  116. // O(log n)  --- max_n = 10^12 ~ 10^15    --- Function to find the difference of two numbers under modulo
  117. ll Mod_sub(ll a, ll b) {
  118.     return ((a % MOD) - (b % MOD) + MOD) % MOD;
  119. }
  120. //---------------------------------------------------------------------------------------------
  121. // O(log n)  --- max_n = 10^12 ~ 10^15    --- Function to find the product of two numbers under modulo
  122. ll Mod_mul(ll a, ll b) {
  123.     return ((a % MOD) * (b % MOD)) % MOD;
  124. }
  125. //---------------------------------------------------------------------------------------------
  126. // O(log n)  --- max_n = 10^12 ~ 10^15    --- recursive function to calculate (n^pow) % MOD 
  127. ll Fast_power_Mod(ll n, ll pow) { 
  128. 	if (pow == 0)return 1;       // base case
  129. 	ll val = Fast_power_Mod(n, pow / 2);
  130. 	ll res = val * val % MOD;
  131.     if (pow % 2 != 0) {
  132.         res *= n;
  133. 		res %= MOD;
  134.     }
  135. 	return res;
  136. }
  137. //----------------------------------------------------------------------------------------------
  138. // O(log n)  --- max_n = 10^12 ~ 10^15    --- Function to convert a number to a different base
  139. vector<int>Convert(ll n, ll base) {
  140.         vector<int>digits;
  141.         while (n) {
  142.             digits.push_back(n % base);
  143.             n /= base;
  144. 		}
  145. 		reverse(all(digits));
  146.         return digits;
  147. }
  148. //----------------------------------------------------------------------------------------------
  149. ll clearBit(ll n, ll idx) {
  150.     return (n & (~(1LL << idx)));
  151. }
  152. //----------------------------------------------------------------------------------------------
  153. ll setBit(ll n, ll idx) {
  154.     return (n | (1LL << idx));
  155. }
  156. //----------------------------------------------------------------------------------------------
  157. ll toggleBit(ll n, ll idx) {
  158.     return (n ^ (1LL << idx));
  159. }
  160. //----------------------------------------------------------------------------------------------
  161. bool is_odd(ll n) {
  162. 	return (n & 1);
  163. }
  164. //----------------------------------------------------------------------------------------------
  165. bool knowBit(ll n,ll idx) {
  166.     return ((n>>idx)&1);
  167. }
  168. //----------------------------------------------------------------------------------------------
  169. int count_ones(ll x)
  170. {
  171.     int cnt = 0;
  172.     while (x)
  173.     {
  174.         x &= (x - 1);
  175.         cnt++;
  176.     }
  177.     return cnt;
  178. }
  179. //--------------------------------------------------------------------------
  180. bool isPowerof22(ll n) { // مش هتشتغل هنا هنعمل فانكشن غيرها
  181. 	return (__builtin_popcountll(n) == 1);    // count 1 bits       use fun popcount me
  182. }
  183. bool isPowerof2(ll n)
  184. {
  185.     return n > 0 && (n & (n - 1)) == 0;
  186. }
  187.  
  188. void solve() {
  189.    // sieve_of_divisors();
  190.     cout << LCM(36, 40);
  191.  
  192. }
  193. int main() {
  194.     ELMOO();
  195.     ll t = 1;
  196.     // cin >> t;
  197.     while (t--) {
  198.         solve();
  199.     }
  200.     return 0;
  201. }
Success #stdin #stdout 0.03s 121432KB
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https://ideone.com/yB7utf
language:
C++ (gcc 8.3)
created:
13 hours ago
visibility:
public

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