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// 3539. Find Sum of Array Product of Magical Sequences
#include "leetcode.h"
/*
* you are given two integers 'm' and 'k', and an integer array 'nums'. a
* sequence of integers 'seq' is called magical if 'seq' has a size of 'm', '0
* <= seq[i] < nums.len', and the binary representation of '2^seq[0] + 2^seq[1]
* + ...' has 'k' set bits. the array product of this sequence is defined as
* 'prod(seq) = (nums[seq[0]] * nums[seq[1]] * ...)'. return the sum of the
* array products for all valid magical sequences. since the answer may be very
* large, return it modulo 10^9+7. a set bit refers to a bit in the binary
* represenation of a number that has a value of 1
*/
static int mod = 1e9 + 7;
int magicalSum(int m, int k, int *nums, int numsSize) {
int **pow_tab = (int **)malloc(numsSize * sizeof(int *));
for (int i = 0; i < numsSize; i++) {
pow_tab[i] = (int *)malloc((m + 1) * sizeof(int));
pow_tab[i][0] = 1;
for (int j = 1; j <= m; j++) {
long long val = (1LL * pow_tab[i][j - 1] * nums[i]) % mod;
pow_tab[i][j] = (int)val;
}
}
int **cmb = (int **)malloc((m + 1) * sizeof(int *));
for (int i = 0; i <= m; i++) {
cmb[i] = (int *)malloc((m + 1) * sizeof(int));
memset(cmb[i], 0, (m + 1) * sizeof(int));
}
for (int i = 0; i <= m; i++) {
cmb[i][0] = 1;
for (int j = 1; j <= i; j++) {
cmb[i][j] = cmb[i - 1][j - 1] + cmb[i - 1][j];
if (cmb[i][j] >= mod)
cmb[i][j] -= mod;
}
}
int ***dp = (int ***)malloc((m + 1) * sizeof(int **));
int ***next = (int ***)malloc((m + 1) * sizeof(int **));
for (int i = 0; i <= m; i++) {
dp[i] = (int **)malloc((m + 1) * sizeof(int *));
next[i] = (int **)malloc((m + 1) * sizeof(int *));
for (int j = 0; j <= m; j++) {
dp[i][j] = (int *)calloc(k + 1, sizeof(int));
next[i][j] = (int *)calloc(k + 1, sizeof(int));
}
}
dp[m][0][0] = 1;
for (int i = 0; i < numsSize; i++) {
for (int j = 0; j <= m; j++)
for (int c = 0; c <= m; c++)
memset(next[j][c], 0, (k + 1) * sizeof(int));
for (int rem = 0; rem <= m; rem++) {
for (int carry = 0; carry <= m; carry++)
for (int ones = 0; ones <= k; ones++) {
int base = dp[rem][carry][ones];
if (!base)
continue;
for (int c = 0; c <= rem; c++) {
int tmp = c + carry;
int bit = tmp & 1;
int ones2 = ones + bit;
if (ones2 > k)
continue;
int carry2 = tmp >> 1;
int rem2 = rem - c;
long long add = base;
add = (add * cmb[rem][c]) % mod;
add = (add * pow_tab[i][c]) % mod;
int curr = next[rem2][carry2][ones2] + (int)add;
if (curr >= mod)
curr -= mod;
next[rem2][carry2][ones2] = curr;
}
}
}
int ***tmp = dp;
dp = next;
next = tmp;
}
long long ans = 0;
for (int carry = 0; carry <= m; carry++) {
int extra = __builtin_popcount(carry);
int need = k - extra;
if (need >= 0 && need <= k) {
ans += dp[0][carry][need];
if (ans >= mod)
ans -= mod;
}
}
for (int i = 0; i < numsSize; i++)
free(pow_tab[i]);
free(pow_tab);
for (int i = 0; i <= m; i++)
free(cmb[i]);
free(cmb);
for (int rem = 0; rem <= m; rem++) {
for (int carry = 0; carry <= m; carry++) {
free(dp[rem][carry]);
free(next[rem][carry]);
}
free(dp[rem]);
free(next[rem]);
}
free(dp);
free(next);
return (int)(ans % mod);
}
int main() {
int n1[] = {1, 10, 100, 10000, 1000000};
int n2[] = {5, 4, 3, 2, 1};
int n3[] = {28};
printf("%d\n", magicalSum(5, 5, n1, ARRAY_SIZE(n1))); // expect: 991600007
printf("%d\n", magicalSum(2, 2, n2, ARRAY_SIZE(n2))); // expect: 170
printf("%d\n", magicalSum(1, 1, n3, ARRAY_SIZE(n3))); // expect: 28
}
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