Here you will find solutions of many problems on spoj. If you want solution of some problem which is not listed in blog or have doubt regarding any spoj problem (which i have solved) or any programming concept (data structure) you can mail me @ raj.nishant360@gmail.com
And my humble request to you all that don't copy the code only try to understand the logic and algorithm behind the code. I have started this because if you tried as hard as you can and still can't find any solution to the problem then you can refer to this.
You can read my answer how to start competitive programming CLICK HERE

Tuesday, December 30, 2014

OPSL-Operation Searchlight

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;
}