**To improve your coffee, improve your grind.**

Chances are you’ve heard this advice (or some variation of it) before. And for good reason – it can save you a… 1,402 more words

**To improve your coffee, improve your grind.**

Chances are you’ve heard this advice (or some variation of it) before. And for good reason – it can save you a… 1,402 more words

They went to sea in a sieve they did

In a sieve they went to sea

In spite of all their friends could say

On a winter’s morn on a stormy day… 613 more words

wouldn’t you think

in this day and age

with all our technology

knowledge and monery

that we could make

each and every month

the exact same length? 36 more words

Proverbs 1:8 Hear, my son, your father’s instruction,

and forsake not your mother’s teaching,

9 for they are a graceful garland for your head

and pendants for your neck. 97 more words

For this problem[], the statement is fairly simple.

1,102 more wordsGiven an integer

N (3<=N<=2^32),you will have to find the largest prime number, that is strictly smaller than N.

I have deployed the sieve.

Yes, this is as ominous as it sounds. Shudder, if you please.

In an attempt not to be overtly wordy in my first Cozy Mystery, I have soldered a fine-meshed strainer to my head. 42 more words

/****************************************** Mobarak Hosen Shakil ICE, Islamic University ID: mhiceiuk(all), 29698(LOJ) E-mail: mhiceiuk @ (Gmail/Yahoo/FB) Blog: https://iuconvergent.wordpress.com *******************************************/ #include<bits/stdc++.h> #define sroot 1000007 using namespace std; vector <int> primes; int sieve() { bool a; memset(a, 0, sizeof(a)); for(int i = 2; i <= sroot; i++) { if(a[i] == 0) { for (int j = i + i; j <= sroot; j = j + i) { a[j] = 1; } } } primes.push_back(2); for (int i = 3; i <= sroot; i+=2) { if(a[i] == 0) { primes.push_back(i); } } } int main() { sieve(); int t; long long x, sum, k, temp, n; ///cout<<primes.size()<<endl; scanf("%d", &t); for (int cases = 1; cases <= t; cases++) { scanf("%lld", &x); sum = 1; n = sqrt(x); for(int i = 0; i <= primes.size() - 1 and primes[i] <= n ; i++) { k = 0; if(x < primes[i]) { break; } while(x % primes[i] == 0) { x = x / primes[i]; k++; } sum = sum * (k + 1); } if(x > 1) sum *= 2; ///cout << x << endl; sum = sum - 1; printf("Case %d: %lld\n",cases, sum); } }