AMR15B how to solve this question ??
# | User | Rating |
---|---|---|
1 | ecnerwala | 3649 |
2 | Benq | 3581 |
3 | orzdevinwang | 3570 |
4 | Geothermal | 3569 |
4 | cnnfls_csy | 3569 |
6 | tourist | 3565 |
7 | maroonrk | 3531 |
8 | Radewoosh | 3521 |
9 | Um_nik | 3482 |
10 | jiangly | 3468 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
AMR15B how to solve this question ??
Name |
---|
A simpler problem: Given an array of integers, find sum of minimum elements of each subset.
Solution: Sort the array so that A[0]<=A[1]<=...<=A[n-1]. Answer= sum 2^(n-i+1)*A[i].
From now on by f(A) we will denote this value for an array.
Prime factorize each number and for each prime p<=10^5 store a list where for each a[i] which is a multiple of p the power of p in a[i] is stored in the list. Let M[p] be this list p.
Answer= product pf(M[p]) over all primes from 1 to 10^5
f(M[p]) can get quite large so it's advisable to compute it modulo 10^9+6 (because x109 + 6 = 1 mod 10^9+7.
.