bzoj 4815: [Cqoi2017]小Q的表格 [数论]

4815: [Cqoi2017]小Q的表格

题意：
单点修改，查询前缀正方形和。修改后要求满足条件f(a,b)=f(b,a), b×f(a,a+b)=(a+b)*f(a,b)

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一开始sb了认为一次只会改动两三个格子想了个cdq分治做法...

一次会影响很多格子...

经过观察以及((a,b)=(a,a-b)=(a,a+b))发现，每次修改影响所有((i,j)=(a,b))的点对，并且关系为(f(i,j)=frac{i}{a}frac{j}{b} f(a,b))

我们可以只记录(f(d,d))的值(f(d))，其他值都能得到

然后套路推倒，最后得到

[ans = sum_{d=1}^n f(d) sum_{i=1}^{frac{n}{d}}sum_{j=1}^{frac{n}{d}}ij[(i,j)=1] ]

这里用莫比乌斯反演我只会两个分块，不如直接代入phi

[ans = sum_{d=1}^n f(d) S(frac{n}{d}) \ S(n) = sum_{i=1}^n i sum_{j=1}^n j [(i,j)=1] = sum_{i=1}^n varphi(i)*i^2 ]

维护f值可以用分块，总体复杂度(O(nsqrt{n}))

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 4e6+5, mo = 1e9+7, M = 3e3+5, inv2 = (mo+1)/2;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
    return x*f;
}

bool notp[N]; int p[N/10], phi[N]; ll s[N];
void sieve(int n) {
	phi[1] = 1;
	for(int i=2; i<=n; i++) {
		if(!notp[i]) p[++p[0]] = i, phi[i] = i-1;
		for(int j=1; j <= p[0] && i*p[j] <= n; j++) {
			notp[i*p[j]] = 1;
			if(i%p[j] == 0) {
				phi[i*p[j]] = phi[i] * p[j];
				break;
			}
			phi[i*p[j]] = phi[i] * (p[j] - 1);
		}
	}
	for(int i=1; i<=n; i++) s[i] = (s[i-1] + (ll) i * i %mo * phi[i] %mo) %mo;
}

int gcd(int a, int b) {return !b ? a : gcd(b, a%b);}
ll Pow(ll a, int b) {
	ll ans=1;
	for(; b; b>>=1, a=a*a%mo)
		if(b&1) ans=ans*a%mo;
	return ans;
}

int Q, n, a, b, k; ll x;
namespace B {
	int f[N], add[N], a[N];
	int pos[N], block, m;
	struct _blo{int l, r;} b[N];
	void init() {
		block = sqrt(n); m = (n-1)/block+1;
		for(int i=1; i<=n; i++) pos[i] = (i-1)/block+1, a[i] = (ll)i*i%mo, f[i] = (a[i] + f[i-1]) %mo;
		for(int i=1; i<=m; i++) b[i].l = (i-1)*block+1, b[i].r = i*block;
	}
	int que(int x) { return (f[x] + add[pos[x]]) %mo; }
	void cha(int x, int v) {
		int d = (v - a[x] + mo) %mo, r = b[pos[x]].r; a[x] = v; 
		if(d==0) return;
		for(int i=x; i<=r; i++) f[i] = (f[i] + d) %mo;
		for(int i=pos[x]+1; i<=m; i++) add[i] = (add[i] + d) %mo;
	}
} using B::cha; using B::que;

void solve(int n) {
	int ans = 0, r, last = 0, now;
	for(int i=1; i<=n; i=r+1, last=now) {
		r = n/(n/i); now = que(r); 
		ans = (ans + (ll) (now - last + mo) * s[n/i] %mo) %mo;
	}
	printf("%d
", (ans + mo) %mo);
}

int main() {
	freopen("in", "r", stdin);
	Q=read(); n=read();
	sieve(n);
	B::init();
	for(int i=1; i<=Q; i++) {
		a=read(); b=read(); scanf("%lld", &x); k=read();
		x %= mo;
		int d = gcd(a, b);
		cha(d, (ll) d * d %mo * x %mo * Pow((ll) a * b %mo, mo-2) %mo );
		solve(k);
	}
}