数论-同余与逆元

文章目录

• 同余
• • 一元线性同余方程
• 逆元
• • 逆元与除法取模
• 例题
• • HDU-5976

同余

ll inv(ll a, ll m){
	ll x, y;
	ll gcd = extend_gcd(a, m, x, y);
	return	(x % m + m) % m;//注意负数
}

递推求逆元：

inv[1] = 1;
for (i = 2; i < n; i++)
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;

逆元与除法取模

逆元一重要作用就是求除法的模，首先我们看一下模运算：

例题

HDU-5976

HDU-5976 Detachment

Problem Description In a highly developed alien society, the habitats are almost infinite dimensional space. In the history of this planet,there is an old puzzle. You have a line segment with x units’ length representing one dimension.The line segment can be split into a number of small line segments: a1,a2, … (x= a1+a2+…) assigned to different dimensions. And then, the multidimensional space has been established. Now there are two requirements for this space: 1．Two different small line segments cannot be equal ( ai≠aj when i≠j). 2．Make this multidimensional space size s as large as possible (s= a1∗a2*…).Note that it allows to keep one dimension.That’s to say, the number of ai can be only one. Now can you solve this question and find the maximum size of the space?(For the final number is too large,your answer will be modulo 10^9+7) Input The first line is an integer T,meaning the number of test cases. Then T lines follow. Each line contains one integer x. 1≤T≤10^6, 1≤x≤10^9 Output Maximum s you can get modulo 10^9+7. Note that we wants to be greatest product before modulo 10^9+7. Sample Input 1 4 Sample Output 4

将一个数拆成若干数，求其乘积最大。 先上结论：一个数n拆成m个数使其乘积最大，则拆成m个n/m；如果nm不整除则拆成一段连续自然数(从2开始，剩下的往前摊)；如果不限制m，则拆成最多的3，剩下的拆成2。证明参考 这题要求数不能相同，所以拆成从2开始的连续数，然后就是预处理+逆元取模即可，详见代码。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e4 + 10;
const int mod = 1e9 + 7;
ll sum[maxn];
void init(){
	sum[0] = sum[1] = 1;
	for (int i = 2; i < maxn; i++)
		sum[i] = i * sum[i - 1] % mod;
}
ll extend_gcd(ll a, ll b, ll& x, ll& y){
	if (b == 0){
		x = 1; y = 0;
		return a;
	}
	ll gcd = extend_gcd(b, a % b, y, x);
	y -= x * (a / b);
	return gcd;
}
ll inv(ll a, ll m){
	ll x, y;
	ll gcd = extend_gcd(a, m, x, y);
	return	x = (x % mod + mod) % mod;
}
int main(){
	ll t, s;
	init();
	scanf("%lld", &t);
	while (t--){
		scanf("%lld", &s);
		if (s <= 4) {
			printf("%d\n", s);
			continue;
		}
		ll n = (ll)((-1 + sqrt(8 * s + 9.0)) / 2 + 0.5);
		while ((2 + n) * (n - 1) / 2 > s) n--;
		int tmp = s - (2 + n) * (n - 1) / 2;//差值，
		ll mi = 2 + tmp / (n - 1);//相乘中的最小值
		ll mx = tmp % (n - 1) ? n + tmp / (n - 1) + 1 : n + tmp / (n - 1);//相乘中的最大值
		ll temp = 2 + tmp / (n - 1) + n - 1 - tmp % (n - 1);//相乘中间断的那个数
		ll ans = inv(sum[mi - 1], mod) * sum[mx] % mod;
		if (temp <= mx)
			ans = (ans * inv(temp, mod)) % mod;
		printf("%lld\n", ans);	
	}
	return 0;
}

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