Can anyone help me with this problem:
Problem: Number of Solution of equation x_1 + x_2 + x_3 + ... x_k = N
for all i: x_i>=0 and x_i<=x_(i+1) constraints:[N<= 10^9 k<=10]
[basically non-decreasing solutions]
# | User | Rating |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 161 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | luogu_official | 151 |
Can anyone help me with this problem:
Problem: Number of Solution of equation x_1 + x_2 + x_3 + ... x_k = N
for all i: x_i>=0 and x_i<=x_(i+1) constraints:[N<= 10^9 k<=10]
[basically non-decreasing solutions]
Name |
---|
Generate the solution for small N and K and look for a recurrence relation.
Wouldn't you get a O(NK) solution with that?
By the way, I already thought of an O(NK) solution:
You can get O(logN * K3) by using matrix exponentiation (if your formula allows it).
EDIT: That formula is not compatible with matrix exponentiation. Maybe there is another one.
I think it is compatible because f(n, k - 1) + f(n - k, k) = f(n - k, k) + f(n - k + 1, k - 1) + ... + f(n - 1, 1).
For example when k = 3:
Its time complexity is O(K6logN).
I think this should work...
The 'x' values must be integers?
Aren't there infinitely many solutions if we don't assume it?
Hint 1: Prove that the equation is equivalent to y_1 + 2y_2 + ... + ky_k = N. Hint 2: Digit DP. Represent each y_i as a binary number and fix the parity.