1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
|
// 1799. Maximize Score After N Operations
#include "leetcode.h"
/*
* given 'nums', an array of positive integers of size '2 * n'. you must perform
* 'n' operations on this array. in the i'th operation (1-index), you will:
* - choose two elements 'x' and 'y'
* - receive a score of 'i * gcd(x, y)'
* - remove 'x' and 'y' from 'nums'
* return the maximum score you can receive after performing 'n' operations. the
* function 'gcd(x, y)' is the greatest common divisor.
*/
class Solution {
public:
int maxScore(vector<int> nums) {
m = nums.size();
n = m / 2;
return dp(nums, 1, 0);
}
private:
int gcd_memo[14][14] = {};
int memo[1 << 4] = {};
int m, n;
int dp(vector<int> &nums, int op, int mask) {
if (op > n)
return 0;
if (memo[mask])
return memo[mask];
for (int i = 0; i < m; ++i) {
if ((mask >> i) & 0x1)
continue;
for (int j = i + 1; j < m; ++j) {
if ((mask >> j) & 0x1)
continue;
int new_mask = (0x1 << i) | (0x1 << j) | mask;
int score = op * gcd(nums, i, j) + dp(nums, op + 1, new_mask);
memo[mask] = max(memo[mask], score);
}
}
return memo[mask];
}
int gcd(vector<int> &nums, int i, int j) {
if (gcd_memo[i][j])
return gcd_memo[i][j];
return gcd_memo[i][j] = __gcd(nums[i], nums[j]);
}
};
int main() {
Solution obj;
vector<int> n1 = {1, 2};
vector<int> n2 = {3, 4, 6, 8};
vector<int> n3 = {1, 2, 3, 4, 5, 6};
printf("%d\n", obj.maxScore(n1)); // expect: 1
printf("%d\n", obj.maxScore(n2)); // expect: 11
printf("%d\n", obj.maxScore(n3)); // expect: 14
}
|