CodeForces 986F Oppa Funcan Style Remastered

洛谷传送门

CF 传送门

有意思的。

对 \(k\) 分解质因数，题目实际上是想让我们解一个 \(\sum\limits_{i = 1}^m a_i x_i = n\) 的方程。

考虑 \(m = 1\) 特判，\(m = 2\) exgcd。\(m = 3\) 时发现 \(\min\limits_{i = 1}^m a_i \le k^{\frac{1}{3}} \le 10^5\)，所以可以跑同余最短路。设 \(f_i\) 为能组成的 \(\bmod a_1 = i\) 的最小数。那么就直接判 \(f_{n \bmod a_i} \le n\) 即可。

同余最短路还在写最短路？感觉不如转圈！

code

// Problem: F. Oppa Funcan Style Remastered
// Contest: Codeforces - Codeforces Round 485 (Div. 1)
// URL: https://codeforces.com/problemset/problem/986/F
// Memory Limit: 256 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef __int128 lll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 10050;
const int N = 32000000;

ll n, ans[maxn], m, f[maxn * 10];
int pr[N / 10], tot;
bool vis[N + 5];
struct node {
	ll n, k, i;
} qq[maxn];

inline void init() {
	for (int i = 2; i <= N; ++i) {
		if (!vis[i]) {
			pr[++tot] = i;
		}
		for (int j = 1; j <= tot && i * pr[j] <= N; ++j) {
			vis[i * pr[j]] = 1;
			if (i % pr[j] == 0) {
				break;
			}
		}
	}
}

ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1;
		y = 0;
		return a;
	}
	ll d = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return d;
}

void solve() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld%lld", &qq[i].n, &qq[i].k);
		qq[i].i = i;
	}
	sort(qq + 1, qq + n + 1, [&](const node &a, const node &b) {
		return a.k < b.k;
	});
	for (int i = 1, j = 1; i <= n; i = (++j)) {
		while (j < n && qq[j + 1].k == qq[i].k) {
			++j;
		}
		m = qq[i].k;
		if (m == 1) {
			for (int k = i; k <= j; ++k) {
				ans[qq[k].i] = 0;
			}
			continue;
		}
		vector<ll> P;
		for (int k = 1; k <= tot && 1LL * pr[k] * pr[k] <= m; ++k) {
			if (m % pr[k] == 0) {
				P.pb(pr[k]);
				while (m % pr[k] == 0) {
					m /= pr[k];
				}
			}
		}
		if (m > 1) {
			P.pb(m);
		}
		if ((int)P.size() == 1) {
			for (int k = i; k <= j; ++k) {
				ans[qq[k].i] = (qq[k].n % P[0] == 0);
			}
			continue;
		}
		if ((int)P.size() == 2) {
			ll p, q;
			ll d = exgcd(P[0], P[1], p, q);
			for (int k = i; k <= j; ++k) {
				lll x = p, y = q;
				if (qq[k].n % d) {
					ans[qq[k].i] = 0;
					continue;
				}
				x *= (qq[k].n / d);
				y *= (qq[k].n / d);
				lll pp = P[1] / d, qq = P[0] / d;
				if (x < 0) {
					lll t = (-x + pp - 1) / pp;
					x += t * pp;
					y -= t * qq;
				}
				if (x > 0) {
					lll t = x / pp;
					x -= t * pp;
					y += t * qq;
				}
				ans[::qq[k].i] = (x >= 0 && y >= 0);
			}
			continue;
		}
		for (int i = 1; i <= P[0]; ++i) {
			f[i] = 8e18;
		}
		f[0] = 0;
		ll t = P[0];
		for (int i = 1; i < (int)P.size(); ++i) {
			for (ll j = 0, lim = __gcd(P[i], t); j < lim; ++j) {
				for (ll k = j, c = 0; c < 2; c += (k == j)) {
					ll p = (k + P[i]) % t;
					f[p] = min(f[p], f[k] + P[i]);
					k = p;
				}
			}
		}
		for (int k = i; k <= j; ++k) {
			ans[qq[k].i] = (f[qq[k].n % t] <= qq[k].n);
		}
	}
	for (int i = 1; i <= n; ++i) {
		puts(ans[i] ? "YES" : "NO");
	}
}

int main() {
	init();
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}