YBTOJ 6.3同余问题

A.同余方程

详见扩展欧几里得算法学习笔记

点击查看代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;

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

int main() {
	ll a, b, x, y;
	scanf("%lld%lld", &a, &b);
	exgcd(a, b, x, y);
	x = (x + b) % b;
	printf("%lld",x);
	
	return 0;
}

---

B.约数之和

考虑算术基本定理
求

$\prod _{i=1}^n\sum _{j=0}^{b\times c_i}{p}_i^j$

根据等比数列求和公式 得

$\prod _{i=1}^n(\frac{p_i^{b\times c_i+1}-1}{p_i-1})$

考虑 $p_i-1$ 与 $9901$ 不互质的情况
则有 $p_i-1\equiv 0(\text{mod }9901)$
$p_i\equiv 1(\text{mod }9901)$
那么对于任意 $x$ 都有 $p_i^x\equiv 1(\text{mod }9901)$
那么显然 $\sum _{j=0}^{b\times c_i}{p}_i^j\mathrm{mod}9901=b\times c_i+1$

点击查看代码

#include <bits/stdc++.h>
#define ll long long

using namespace std;

const ll mod = 9901;
ll a, b, ans = 1;
int tot;

ll ksm(ll x, ll y) {
	ll ret = 1;
	while (y) {
		if (y & 1) ret = ret * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return ret;
}

ll inv(ll x) {
	return ksm(x, mod - 2);
}

void work(ll x, ll y) {
	if ((x - 1) % mod == 0) ans = (ans * (b + 1) % mod) % mod;
	else ans = (ans * (ksm(x, y + 1) - 1) % mod * inv(x - 1)) % mod;
}

int main() {
	scanf("%lld%lld", &a, &b);
	for (ll i = 2; i * i <= a; ++i) {
		if (a % i == 0) {
			int num = 0;
			while (a % i == 0) {
				a /= i;
				num++;
			}
			work(i, num * b);
		}
	}
	if (a > 1) work(a, 1 * b);
	printf("%lld",ans);
	
	return 0;
}

---

C.线性求逆元

• 解法1

考虑构造 $mod=\lfloor {\displaystyle \frac{mod}{x}}\rfloor \times x+r$

转化为 $\lfloor {\displaystyle \frac{mod}{x}}\rfloor \times x+r\equiv 0(\text{mod }mod)$

同乘 $x^{-1}\times r^{-1}$ 得
$\lfloor {\displaystyle \frac{mod}{x}}\rfloor \times r^{-1}+x^{-1}\equiv 0(\text{mod }mod)$

所以有
$x^{-1}\equiv -\lfloor {\displaystyle \frac{mod}{x}}\rfloor \times r^{-1}(\text{mod }mod)$

由于 $r=mod\mathrm{mod}x$

显然 $r<x$

也就是说我们已经知道了 $r^{-1}$ 就可以直接求出 $x^{-1}$ 了

inv[1] = 1;
for (int i = 2; i <= n; ++i) {
	inv[i] = ((p - p / i) * inv[p % i] + p) % p;
}

• 解法二

来自 @rabbit_lb 可用来求阶乘的逆元

首先用快速幂求出 $n{!}^{-1}$
那么 $(n-1){!}^{-1}=n{!}^{-1}\times n$
然后一直推下去即可

---

D.中国剩余定理

这章怎么全是板子题

详见中国剩余定理学习笔记

点击查看代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 11;
ll a[N], b[N];
ll pai = 1;
int k;
ll ans;

ll ksc(ll x, ll y) {
	ll ret = 0;
	while (y) {
		if (y & 1) ret = (ret + x) % pai;
		x = (x + x) % pai;
		y >>= 1;
	}
	return ret;
}

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

int main() {
	scanf("%d", &k);
	for (int i = 1; i <= k; ++i) scanf("%lld%lld", &b[i], &a[i]);
	for (int i = 1; i <= k; ++i) pai *= b[i];
	
	for (int i = 1; i <= k; ++i) {
		ll x, y;
		exgcd(pai / b[i], b[i], x, y);
		x = (x % b[i] + b[i]) % b[i];
		a[i] = (a[i] % b[i] + b[i]) % b[i];
		ans = (ans + ksc(pai / b[i] * x, a[i])) % pai;
	}
	
	printf("%lld", ans);
	
	return 0;
}