### Operation Searchlight

Given below c++ code is for OPSL spoj or Operation Searchlight spoj .

This is an easy problem & should be in tutorial class , here you should know sieve and how to calculate LCM of numbers.

#include <bits/stdc++.h> using namespace std; #define MOD 1000000007 #define LL long long int p[1000009]; int prime[100000]; int pre() { for(int i = 3 ; i * i <= 1000000 ; i+=2){ if(!p[i]) for(int j = i*i ; j <= 1000000 ; j+=i) p[j] = 1; } prime[0] = 2; int k = 1; for(int i = 3 ; i <= 1000000 ; i+=2) if(!p[i]) prime[k++] = i; return k; } int main() { int t; int len = pre(); cin>>t; for(int test = 1 ; test <= t ; test++){ int n; scanf("%d",&n); int arr[n+9]; for(int i =0 ; i < n ; i++) scanf("%d",&arr[i]); int a[len+9]; memset(a , 0 , sizeof a); for(int i = 0 ; i < n ; i ++){ LL temp = arr[i]; for(int j = 0 ; (LL)prime[j] * prime[j] <= temp ; j++){ int count = 0; while(temp%prime[j] == 0){ temp/=prime[j]; count++; } a[j] = max(a[j] , count); } if(temp > 1){ int pos = lower_bound(prime , prime + len , temp) - prime; a[pos] = max(a[pos] , 1); } } LL res = 1; for(int i = 0 ; i < len ; i++) { int temp = a[i] ; while(temp--) res = (res * prime[i])%MOD; } printf("Case %d: %lld\n",test ,res); } return 0; }