[SCOI2018]Numazu 的蜜柑 [ 二次剩余 ]

最新推荐文章于 2020-02-19 15:42:07 发布

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最新推荐文章于 2020-02-19 15:42:07 发布

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分类专栏：二次剩余

本文链接：https://blog.youkuaiyun.com/sslz_fsy/article/details/89785571

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博客提及直接解相关情况，引出二次剩余解。若有解，用map在树上进行统计；若无解，则统计0的对数。

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直接解发现 $a_u=\frac{a_v(-A\pm \sqrt{A^2-4B})}{2}$

于是二次剩余解出 $det\equiv\sqrt{A^2-4B}(mod p)$

如果有解, 用map在树上统计一下就可以了, 如果无解, 就统计0的对数

#include<bits/stdc++.h>
#define LL long long
#define N 200050
using namespace std;
LL read(){
	LL cnt = 0, f = 1; char ch = 0;
	while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1;}
	while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
	return cnt * f;
}
int n; LL A, B, p, a[N];
vector<int> v[N];
LL det, w, a1, a2, ans;
LL mul(LL a, LL b){
	LL ans = 0; for(;b;b>>=1){
		if(b&1) ans = (ans+a) % p;
		a = (a+a) % p;
	} return ans;
}
LL power(LL a, LL b){
	LL ans = 1; for(;b;b>>=1){
		if(b&1) ans = mul(ans, a);
		a = mul(a, a);
	} return ans;
}
LL legander(LL x){
	LL a = power(x, (p-1)/2);
	if(a + 1 == p) return -1;
	return a;
}
struct Node{
	LL x, y;
	Node(LL _x=0, LL _y=0){ x = _x, y = _y;}
	Node operator * (const Node &a)
	{ return Node((mul(x, a.x) + mul(mul(y, a.y), w))%p, (mul(x, a.y) + mul(y, a.x))%p); } 
};
Node Power(Node a, LL b){
	Node ans = Node(1, 0);
	for(;b;b>>=1){
		if(b&1) ans = ans * a;
		a = a * a;
	} return ans;
}
LL Solve(LL x){
	LL a;
	while(1){
		a = rand() % p;
		w = (mul(a, a) - x + p) % p;
		if(legander(w) == -1) break;
	}
	Node res = Node(a, 1ll);
	res = Power(res, (p+1) / 2);
	return res.x;
}
map<LL, int> cnt;
void dfs1(int u){
	ans += cnt[a[u]];
	LL v1 = mul(a1, a[u]), v2 = mul(a2, a[u]);
	(v1 == v2) ? cnt[v1]++ : (cnt[v1]++, cnt[v2]++);
	for(int i=0; i<v[u].size(); i++) dfs1(v[u][i]);
	(v1 == v2) ? cnt[v1]-- : (cnt[v1]--, cnt[v2]--);
}
LL num;
void dfs2(int u){
	if(a[u] == 0) ans += num, num++;
	for(int i=0; i<v[u].size(); i++) dfs2(v[u][i]);
	if(a[u] == 0) num--;
}
int main(){
	n = read(), p = read(), A = read(), B = read();
	for(int i=1; i<=n; i++) a[i] = read();
	for(int i=2; i<=n; i++) v[read()].push_back(i);
	det = (mul(A, A) - (B * 4) % p + p) % p;
	if(legander(det) != -1){
		det = Solve(det);
		a1 = mul(((det - A + p) % p), power(2, p-2));
		a2 = mul(((-det - A + p + p) % p), power(2, p-2));
		dfs1(1);
	}
	else dfs2(1);
	printf("%lld", ans);
	return 0;
}