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5 months ago+ 0 comments

When finding rotations of a number, don't forget about zeroes in the middle. You should rotate honestly including zeroes (example for a not prime number):

104080 -> 040801, 408010, 080104, 801040, 010408

Use memoization in the isprime() function. You will have overhead otherwise.

8 months ago+ 0 comments

Use sieve of eratosthenes, find whether the number and their rotations are prime and add them to a set and find the sum. I have the code in Python if you don't understand~

2 years ago+ 0 comments

Hey! I was stucked in this problem and so frustrated So, here is the solution (in c++)

int prime(int x){ for(int i=2;i*i<=x;i++){ if(x%i == 0) return false; } return true; } int p(int x){ int d = 1; int digs = 0; while(d <= x){ d *= 10; digs++; } d/=10; for(int k=0;k

                    THANKS ME LATER!
        //cout << i <<endl;
    }
}
printf("%d\n", sum);

2 years ago+ 0 comments

Python O(1) sol:

n = int(input())
e = sorted([2,3,5,7,11,13,31,37,17,71,73,79,97,113,131,311,197,971,719,199,991,919,337,373,733,1193,1931,9311,3119,3779,7793,7937,9377,11939,19391,93911,39119,91193,19937,99371,93719,37199,71993,193939,939391,393919,939193,391939,919393,199933,999331,993319,933199,331999,319993,])
res = 0
for i in e:
    if i <= n:
        res += i
    else:
        break
print(res)

2 years ago+ 0 comments

C++ Solution :

#include <bits/stdc++.h>
using namespace std;
#define all(v) (v).begin(), (v).end()
#define debug(x) cout << #x << " = " << x << endl
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
inline ll Mod(ll x, ll mod) { return x % mod >= 0 ? x % mod : x % mod + mod; }

bool *is_prime = new bool[1000001];

void sieve(int N)
{

    for (int i = 0; i <= N; i++)
        is_prime[i] = 1;

    is_prime[0] = 0;
    is_prime[1] = 0;
    for (int i = 2; i * i <= N; ++i)
    {
        if (is_prime[i])
        {
            for (int j = i * i; j <= N; j += i)
                is_prime[j] = 0;
        }
    }
}

bool is_rotate(int n, int p)
{
    for (int i = 0; i < 7; i++)
    {
        int y = n % 10;
        n /= 10;
        n += y * p;
        if (!is_prime[n])
            return 0;
    }
    return 1;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    sieve(1000000);
    ll *ans = new ll[1000001];
    memset(ans, 0, sizeof(ans));
    int p = 1;
    for (int i = 2; i < 1000001; i++)
    {
        if (i == p * 10)
            p *= 10;
        if (is_prime[i] == 0 || is_rotate(i, p) == 0)
        {
            ans[i] = ans[i - 1];
            continue;
        }
        ans[i] = ans[i - 1] + i;
    }
    int n;
    cin >> n;
    cout << ans[n];
}

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