Find the value of polynomial at some K different integers, modulo 786433 - c++

Question

You are given a polynomial of degree N with integer coefficients. Your task is to find the value of this polynomial at some K different integers, modulo 786433.
Input
The first line of the input contains an integer N denoting the degree of the polynomial.
The following line of each test case contains (N+1) integers denoting the coefficients of the polynomial. The ith numbers in this line denotes the coefficient a_(i-1) in the polynomial a_0 + a_1 × x_1 + a_2 × x_2 + ... + a_N × x_N.
The following line contains a single integer Q denoting the number of queries.
The jth of the following Q lines contains an integer number x_j denoting the query.
Output
For each query, output a single line containing the answer to the corresponding query. In other words, the jth line of the output should have an integer equal to a_0 + a_1 × x_j + a_2 × x_j^2 + ... + a_N × x_j^N modulo 786433.
Constraints and Subtasks
0 ≤ a_i, x_j < 786433
Subtask #1 (37 points): 0 ≤ N, Q ≤ 1000
Subtask #2 (63 points): 0 ≤ N, Q ≤ 2.5 × 10^5
Example
Input:
2
1 2 3
3
7
8
9
Output:
162
209
262
Explanation
Example case 1.
Query 1: 1 + 2 × 7 + 3 × 7 × 7 = 162
Query 2: 1 + 2 × 8 + 3 × 8 × 8 = 209
Query 3: 1 + 2 × 9 + 3 × 9 × 9 = 262
Here is the code that runs in O(n log n) time. I use Fast Fourier Transform to multiply the two polynomials.
Why do I need to use the modulo 786433 operator for all calculations? I understand that this might relate to int overflow? It this normal in competitive programming questions?
#include <cstdio>
#include <algorithm>
#include <vector>
#include <sstream>
#include <iostream>
using namespace std;
#define all(a) (a).begin(),(a).end()
#define pb push_back
#define sz(a) ((int)(a).size())
#define mp make_pair
#define fi first
#define se second
typedef pair<int, int> pint;
typedef long long ll;
typedef vector<int> vi;
#define MOD 786433
#define MAGIC (3*(1<<18))
const int root = 10;
void fft(vi &a, int wn = root)
{
int n = sz(a);
if (n == 3)
{
int a1 = a[0] + a[1] + a[2];
int a2 = (a[0] + a[1] * 1LL * root + a[2] * (root * 1LL * root)) % MOD;
a[1] = a1;
a[2] = a2;
return;
}
vi a0(n / 2), a1(n / 2);
for (int i = 0, j = 0; i<n; i += 2, ++j)
{
a0[j] = a[i];
a1[j] = a[i + 1];
}
int wnp = (wn * 1LL * wn) % MOD;
fft(a0, wnp);
fft(a1, wnp);
int w = 1;
for (int i = 0; i<n / 2; ++i) {
int twiddle = (w * 1LL * a1[i]) % MOD;
a[i] = (a0[i] + twiddle) % MOD;
a[i + n / 2] = (a0[i] - twiddle + MOD) % MOD;
w = (w * 1LL * wn) % MOD;
}
}
int n;
vi coef;
void poly(stringstream& ss)
{
ss >> n;
n++;
for (int i = 0; i<n; i++)
{
int x;
ss >> x;
coef.pb(x);
}
while (sz(coef)<MAGIC)
coef.pb(0);
vi ntt = coef;
fft(ntt);
vector<pint> sm;
sm.pb(mp(0, coef[0]));
int pr = 1;
for (int i = 0; i<sz(ntt); i++)
{
sm.pb(mp(pr, ntt[i]));
pr = (pr * 1LL * root) % MOD;
}
sort(all(sm));
int q;
ss >> q;
while (q--)
{
int x;
ss >> x;
int lo = 0, hi = sz(sm) - 1;
while (lo<hi)
{
int m = (lo + hi) / 2;
if (sm[m].fi<x)
lo = m + 1;
else
hi = m;
}
printf("%d\n", sm[lo].se);
}
}
void test1()
{
stringstream ss;
{
int degree = 2;
ss << degree << "\n";
string coefficients{ "1 2 3" };
ss << coefficients << "\n";
int NoQueries = 3;
ss << NoQueries << "\n";
int query = 7;
ss << query << "\n";
query = 8;
ss << query << "\n";
query = 9;
ss << query << "\n";
}
poly(ss);
}
int main()
{
test1();
return 0;
}
BTW.: This question is from the July 2016 challenge # code chef

Answer 1

Re
” Why do I need to use the modulo 786433 operator for all calculations? I understand that this might relate to int overflow? It this normal in competitive programming questions?
Yes, it's to avoid overflow.
And yes, it's normal in programming problem sets and competitions.