题解 Function Query

传送门

打表发现 \(f(i, j)=f(\frac{i}{\gcd(i, j)}, \frac{j}{\gcd(i, j)})\)
然后发现当 \(x+y=2^k\) 时 \(f(x, y)=k\)
那么枚举 \(i\)，枚举 \(gcd\)，枚举 \(t\)，树状数组修改
复杂度均摊下来是 \(O(n\log^2 n)\) 的

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define fir first
#define sec second
#define pb push_back
#define ll long long
// #define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, q;
int a[N];

namespace force{
	ll ans;
	map<pair<ll, ll>, bool> mp;
	ll f(ll x, ll y) {
		// cout<<"f: "<<x<<' '<<y<<endl;
		if (mp.find({x, y})!=mp.end()) return -1;
		mp[{x, y}]=1;
		if (x==y) return 1;
		else {
			ll t=f(x+y-abs(x-y), abs(x-y));
			if (t==-1) return -1;
			else return t+1;
		}
	}
	void solve() {
		for (int i=1,l,r; i<=q; ++i) {
			l=read(); r=read(); ans=0;
			for (int j=l; j<=r; ++j)
				for (int k=j+1; k<=r; ++k) {
					mp.clear();
					int t=f(a[j], a[k]);
					if (t!=-1) ans+=t; //, cout<<j<<' '<<k<<' '<<t<<endl;
				}
			printf("%lld\n", ans);
		}
	}
}

namespace task1{
	vector<int> div[N];
	ll ans[N], val[N], mp[N];
	vector<pair<int, int>> que[N];
	void upd(int l, int r, int dat) {for (int i=l; i<=r; ++i) val[i]+=dat;}
	ll query(int l, int r) {ll ans=0; for (int i=l; i<=r; ++i) ans+=val[i]; return ans;}
	void solve() {
		for (int i=1,l,r; i<=q; ++i) {
			l=read(); r=read();
			que[l].pb({r, i});
		}
		for (int i=1; i<=n; ++i) mp[a[i]]=i;
		for (int i=1; i<=n; ++i)
			for (int j=i; j<=n; j+=i)
				div[j].pb(i);
		for (int i=n,x,y; i; --i) {
			// cout<<"i: "<<i<<' '<<a[i]<<endl;
			for (auto it:div[a[i]]) {
				x=a[i]/it;
				for (int t=1; t<=17; ++t) {
					y=(1<<t)-x;
					if (y>=1&&y<=n&&__gcd(x, y)==1) {
						y*=it;
						if (y>=1&&y<=n && mp[y]>i) {
							// cout<<a[i]<<' '<<y<<' '<<t<<endl;
							upd(mp[y], mp[y], t);
						}
					}
				}
			}
			for (auto it:que[i]) ans[it.sec]+=query(i, it.fir);
		}
		for (int i=1; i<=q; ++i) printf("%lld\n", ans[i]);
	}
}

namespace task2{
	int mp[N];
	ll ans[N], bit[N];
	vector<int> div[N];
	vector<pair<int, int>> que[N];
	inline void upd(int i, int dat) {for (; i<=n; i+=i&-i) bit[i]+=dat;}
	inline ll query(int i) {ll ans=0; for (; i; i-=i&-i) ans+=bit[i]; return ans;}
	inline int gcd(int a, int b) {return !b?a:gcd(b, a%b);}
	void solve() {
		for (int i=1,l,r; i<=q; ++i) {
			l=read(); r=read();
			que[l].pb({r, i});
		}
		for (int i=1; i<=n; ++i) mp[a[i]]=i;
		for (int i=1; i<=n; ++i)
			for (int j=i; j<=n; j+=i)
				div[j].pb(i);
		ll x, y;
		for (int i=n; i; --i) {
			// cout<<"i: "<<i<<' '<<a[i]<<endl;
			for (auto it:div[a[i]]) {
				x=a[i]/it;
				for (int t=1; t<=17; ++t) {
					y=(1<<t)-x;
					if (y>=1&&y<=n&&gcd(x, y)==1) {
						y*=it;
						if (y>=1&&y<=n && mp[y]>i) {
							// cout<<a[i]<<' '<<y<<' '<<t<<endl;
							upd(mp[y], t);
						}
					}
				}
			}
			for (auto it:que[i]) ans[it.sec]+=query(it.fir);
		}
		for (int i=1; i<=q; ++i) printf("%lld\n", ans[i]);
	}
}

signed main()
{
	freopen("func.in", "r", stdin);
	freopen("func.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	q=read();
	// force::solve();
	// task1::solve();
	task2::solve();

	return 0;
}