Product modulo random numbers - c++

Question

A prime p is fixed. A sequence of n numbers is given, each from 1 to p - 1. It is known that the numbers in the sequence are chosen randomly, equally likely and independently from each other. Choose some numbers from the sequence so that their product, taken modulo p, is equal to the given number x. If no numbers are selected, the product is considered equal to one.
Input:
The first line contains three integers separated by spaces: the length of the sequence n, the prime number p and the desired value x
(n=100, 2<=p<=10^9, 0<x<p)
Next, n integers are written, separated by spaces or line breaks: the sequence a1, a2,. . ., an
(0 <ai <p)
Output:
Print the numbers from the sequence whose product modulo p is equal to x. The order in which numbers are displayed is not important. If there are several possible answers, print any of them
Example:
INPUT:
100 11 4
9 6 1 1 10 4 9 10 3 1 10 1 6 8 3 3 9 8
10 3 7 7 1 3 3 1 5 2 10 4 1 5 6 7 2 6
2 8 3 3 6 7 6 3 1 5 10 2 2 10 9 6 8 6
2 10 3 2 7 4 3 2 8 6 4 1 7 2 10 8 4 9
7 9 8 7 4 7 3 2 8 2 3 7 1 5 2 10 7 1 8
6 4 10 10 3 6 10 2 1
OUTPUT:
4 6 10 9
My solution:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
int n,p,x,y,m,k,tmp;
vector<int> v;
cin >> n >> p >> x;
for (int i = 0; i<n; i++){
cin >> tmp;
v.push_back(tmp);
}
sort(v.begin(), v.end());
v.erase(v.begin(), upper_bound(v.begin(), v.end(), 1));
k=-1;
while(1){
k++;
m = 1;
y = x+p*k;;
vector<int> res;
for (int i = 0; i<n; i++){
if (y == 1) break;
if ( y%v[i] == 0){
res.push_back(v[i]);
m*=v[i];
m%=p;
y = y/v[i];
}
}
if (m==x) {
for (int i = 0; i<res.size(); i++){
cout << res[i] << " ";
}
break;
}
}
return 0;
}
In my solution, I used condition (y=x+k*p, where y is the product of numbers in the answer, and k is some kind of natural number). And also iterated over the value k.
This solution sometimes goes beyond the allotted time. Please tell me a more correct algorithm.

Answer 1

I would consider a backtracking routine over the hashed multiset of the input list. Since p is a prime, at any point we can consider if the current multiple, m, has (multiplicative_inverse(m, p) * x) % p in our multiset (https://en.wikipedia.org/wiki/Multiplicative_inverse). If it exists, we're done. Otherwise, try multiplying either by the same number we are currently visiting in the multiset, or by the next one (keep the result of the multiplication modulo p).
Please see comment below for a link to example code in Python. The example you gave has trivial solutions so it would be helpful to have some non-trivial, as well as challenging examples to test and refine on. Please also clarify if more than one number is expected in the output.

Answer 2

You can use dynamic programming approach. It requires O(p) memory cells and O(p*n) loop iterations. There is possible several optimization (to exclude processing input duplicates, or print longest/shortest selection chain). Following is simplest and basic DP-program, demonstrating this approach.
#include <stdio.h>
#include <stdlib.h>
int data[] = {
9, 6, 1, 1, 10, 4, 9, 10, 3, 1, 10, 1, 6, 8, 3, 3, 9, 8,
10, 3, 7, 7, 1, 3, 3, 1, 5, 2, 10, 4, 1, 5, 6, 7, 2, 6,
2, 8, 3, 3, 6, 7, 6, 3, 1, 5, 10, 2, 2, 10, 9, 6, 8, 6,
2, 10, 3, 2, 7, 4, 3, 2, 8, 6, 4, 1, 7, 2, 10, 8, 4, 9,
7, 9, 8, 7, 4, 7, 3, 2, 8, 2, 3, 7, 1, 5, 2, 10, 7, 1, 8,
6, 4, 10, 10, 3, 6, 10, 2, 1
};
struct elm {
int val; // Value
int prev; // from which elemet we come to this
int n; // add loop cound for prevent multiple use same val
};
void printsol(int n, int p, int x, const int *in) {
struct elm *dp = (struct elm *)calloc(p, sizeof(struct elm));
int i, j;
for(i = 0; i < n; i++) // add initial elements into DP array
dp[in[i]].val = in[i];
for(i = 0; i < n; i++) { // add elements, one by one, to DP array
if(dp[in[i]].val <= 1) // skip secondary "1" multipliers
continue;
for(j = 1; j < p; j++)
if(dp[j].val != 0 && dp[j].n < i) {
int y = ((long)j * in[i]) % p;
dp[y].val = in[i]; // current value, for printout
dp[y].prev = j; // reference to prev element
dp[y].n = n; // loop num, for prevent double reuse
if(x == y && dp[x].n > 0) {
// targed reached - print result, by iterate linklist
int mul = 1;
while(x != 0) {
printf(" %d ", dp[x].val);
mul *= dp[x].val; mul %= p;
x = dp[x].prev;
}
printf("; mul=%d\n", mul);
free(dp);
return;
}
} // for+if
} // for i
free(dp);
}
int main(int argc, char **argv) {
printsol(100, 11, 4, data);
return 0;
}