Given two integers N and M, count the number of integers x between 2 and N! , having the property that all prime factors of x are greater than M. Where 1≤M≤N<10000001 and (N-M)≤100000. Can anyone help me with the logic?
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
Given two integers N and M, count the number of integers x between 2 and N! , having the property that all prime factors of x are greater than M. Where 1≤M≤N<10000001 and (N-M)≤100000. Can anyone help me with the logic?
Name |
---|
Use "sieve of eratosthenes" and find prime numbers that is greater than M and smaller than N all the numbers we want is multiplication of these prime number.
Well but i think it's not possible to find the prime numbers till N!.(N factorial)
I think you can calculate instead the count of numbers that have prime factors smaller than M in the range 2 to N! and substract it from N! - 1. Now let the primes smaller than M pe p1, ..., pk, then by simple inclusion, exclusion the quantitiy you want is equal to which is similar to euler's toitient function formula and you can write it as .
I hope this is correct.
can you post the link of this problem?
I think there's no way to solve this problem right now. Since U need to find all the prime numbered from M to N!. Let the count be X, our answer is 2^X -1.
Here is the problem link : https://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=2435