PROBLEM LINK:
In case the contest did not have two separate pages for Div-1 and Div-2, you can delete one of these above lines.
Author: Nachiket Kanore
Tester: Ajinkya Kharosekar
Editorialist: Ajinkya Kharosekar
DIFFICULTY:
MEDIUM - HARD.
PROBLEM:
You are given 5 arrays: A B C D E each of size N.
Let P(S) be the probability that S=A[u]+B[v]+C[w]+D[x]+E[y] for all 1≤u,v,w,x,y≤N.
You are given Q different queries each with some SS:
- You need to find the corresponding probability P(S) under modulo 998244353998244353
EXPLANATION:
Consider a polynomial for array A, a(x) = sigma (cnt[y].x^y)
where cnt[y] is number of occurences of y in A
Consider 5 different such polynomials for each array
Let P(x) be the product of all these polynomials
Here every coefficient z.x^y represents
z = number of ways to achieve sum of y = sum of single element taken from each array
Find probability:
number of ways / total ways
total ways = n^5
Read more about it here:
https://cp-algorithms.com/algebra/fft.html#toc-tgt-9
SOLUTIONS:
Setter's Solution
/*
Author: Nachiket Kanore
Created: Thursday 28 January 2021 10:30:09 AM IST
(�) I do not believe in a fate that falls on men however they act; but I do believe in a fate that falls on them unless they act.
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <array>
#define int long long
#define pb push_back
#define ALL(x) (x).begin(),(x).end()
#define sz(x) (int)(x.size())
#define FOR(i,L,R) for(int i = (L); i <= (R); i++)
using namespace std;
template<class T> bool cmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool cmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
const int MOD = 998244353;
int mod_pow(int a, int b, int mod = MOD){
int ret = 1;
a %= mod;
while(b > 0){
if(b % 2)
ret = ret * a % mod;
b /= 2;
a = a * a % mod;
}
return ret;
}
int mod_mul(int a, int b) { return (a * b) % MOD; }
int mod_sub(int a, int b) { return (a - b + MOD) % MOD; }
int mod_add(int a, int b) { return (a + b) % MOD; }
int mod_inv(int a) { return mod_pow(a, MOD-2, MOD); }
// NTT template from neal
namespace NTT {
vector<int> roots = {0, 1};
vector<int> bit_reverse;
int max_size = -1;
int root;
bool is_power_of_two(int n) {
return (n & (n - 1)) == 0;
}
int round_up_power_two(int n) {
assert(n > 0);
while (n & (n - 1))
n = (n | (n - 1)) + 1;
return n;
}
// Given n (a power of two), finds k such that n == 1 << k.
int get_length(int n) {
assert(is_power_of_two(n));
return __builtin_ctz(n);
}
// Rearranges the indices to be sorted by lowest bit first, then second lowest, etc., rather than highest bit first.
// This makes even-odd div-conquer much easier.
void bit_reorder(int n, vector<int> &values) {
if ((int) bit_reverse.size() != n) {
bit_reverse.assign(n, 0);
int length = get_length(n);
for (int i = 0; i < n; i++)
bit_reverse[i] = (bit_reverse[i >> 1] >> 1) + ((i & 1) << (length - 1));
}
for (int i = 0; i < n; i++)
if (i < bit_reverse[i])
swap(values[i], values[bit_reverse[i]]);
}
void find_root() {
int order = MOD - 1;
max_size = 1;
while (order % 2 == 0) {
order /= 2;
max_size *= 2;
}
root = 2;
// Find a max_size-th primitive root of MOD.
while (!(mod_pow(root, max_size) == 1 && mod_pow(root, max_size / 2) != 1))
root++;
}
void prepare_roots(int n) {
if (max_size < 0)
find_root();
if ((int) roots.size() >= n)
return;
int length = get_length(roots.size());
roots.resize(n);
// The roots array is set up such that for a given power of two n >= 2, roots[n / 2] through roots[n - 1] are
// the first half of the n-th primitive roots of MOD.
while (1 << length < n) {
// z is a 2^(length + 1)-th primitive root of MOD.
int z = mod_pow(root, max_size >> (length + 1));
for (int i = 1 << (length - 1); i < 1 << length; i++) {
roots[2 * i] = roots[i];
roots[2 * i + 1] = mod_mul(roots[i], z);
}
length++;
}
}
void fft_iterative(int N, vector<int> &values) {
assert(is_power_of_two(N));
prepare_roots(N);
bit_reorder(N, values);
assert(N <= max_size);
for (int n = 1; n < N; n *= 2)
for (int start = 0; start < N; start += 2 * n)
for (int i = 0; i < n; i++) {
int even = values[start + i];
int odd = mod_mul(values[start + n + i], roots[n + i]);
values[start + n + i] = mod_sub(even, odd);
values[start + i] = mod_add(even, odd);
}
}
const int FFT_CUTOFF = 150;
vector<int> mod_multiply(vector<int> left, vector<int> right, bool trim = true) {
int n = left.size();
int m = right.size();
for (int i = 0; i < n; i++)
assert(0 <= left[i] && left[i] < MOD);
for (int i = 0; i < m; i++)
assert(0 <= right[i] && right[i] < MOD);
// Brute force when either n or m is small enough.
if (min(n, m) < FFT_CUTOFF) {
const uint64_t ULL_BOUND = numeric_limits<uint64_t>::max() - (uint64_t) MOD * MOD;
vector<uint64_t> result(n + m - 1);
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
result[i + j] += (uint64_t) left[i] * right[j];
if (result[i + j] > ULL_BOUND)
result[i + j] %= MOD;
}
for (int i = 0; i < (int) result.size(); i++)
if (result[i] >= (uint64_t) MOD)
result[i] %= MOD;
if (trim) {
while (!result.empty() && result.back() == 0)
result.pop_back();
}
return vector<int>(result.begin(), result.end());
}
int N = round_up_power_two(n + m - 1);
left.resize(N);
right.resize(N);
bool equal = left == right;
fft_iterative(N, left);
if (equal)
right = left;
else
fft_iterative(N, right);
int inv_N = mod_inv(N);
for (int i = 0; i < N; i++)
left[i] = mod_mul(mod_mul(left[i], right[i]), inv_N);
reverse(left.begin() + 1, left.end());
fft_iterative(N, left);
left.resize(n + m - 1);
if (trim) {
while (!left.empty() && left.back() == 0)
left.pop_back();
}
return left;
}
vector<int> power(const vector<int> &v, int exponent) {
assert(exponent >= 0);
vector<int> result = {1};
if (exponent == 0)
return result;
for (int k = 31 - __builtin_clz(exponent); k >= 0; k--) {
result = mod_multiply(result, result);
if (exponent & 1 << k)
result = mod_multiply(result, v);
}
return result;
}
};
int32_t main() {
ios::sync_with_stdio(0); cin.tie(0);
/*
int n; cin >> n;
n >>= 1;
int k;
cin >> k;
vector<int> poly(10);
while(k--) {
int u;
cin >> u;
poly[u] = 1;
}
vector<int> ans_poly = NTT::power(poly, n);
int ans = 0;
for (int x : ans_poly) ans = (ans + x * x) % MOD;
cout << ans << "\n";
*/
int n; cin >> n;
vector<int> A(n), B(n), C(n), D(n), E(n);
for (int &x : A) cin >> x;
sort(ALL(A));
for (int &x : B) cin >> x;
sort(ALL(B));
for (int &x : C) cin >> x;
sort(ALL(C));
for (int &x : D) cin >> x;
sort(ALL(D));
for (int &x : E) cin >> x;
sort(ALL(E));
vector<int> a(A.back() + 1); for (int &x : A) a[x]++;
vector<int> b(B.back() + 1); for (int &x : B) b[x]++;
vector<int> c(C.back() + 1); for (int &x : C) c[x]++;
vector<int> d(D.back() + 1); for (int &x : D) d[x]++;
vector<int> e(E.back() + 1); for (int &x : E) e[x]++;
vector<int> result = NTT::mod_multiply(a, b);
result = NTT::mod_multiply(result, c);
result = NTT::mod_multiply(result, d);
result = NTT::mod_multiply(result, e);
/*
map<int,int> brute;
for (int one : A)
for (int two : B)
for (int three : C)
for (int four : D)
for (int five : E)
brute[one + two + three + four + five]++;
for (int i = 0; i < sz(result); i++) {
assert(result[i] == brute[i]);
// cerr << i << " : " << result[i] << '\n';
}
*/
int denominator = mod_pow(n, 5, MOD);
denominator = mod_inv(denominator);
int Q; cin >> Q;
while (Q--) {
int sum; cin >> sum;
int numerator = (sum < sz(result) ? result[sum] : 0);
cout << mod_mul(numerator, denominator) << '\n';
// int my = mod_mul(numerator, denominator);
// int br = mod_mul(brute[sum], mod_inv(mod_pow(n, 5, MOD)));
// assert(my == br);
// cout << my << ' ';
}
}