Hdu 5726 （st+rmq+二分 求相同gcd的区间数量，hash_table计数）

感谢学长的板子

$Orz$：https://blog.csdn.net/j2_o2/article/details/102677710

自己演了自己一天

参考代码

#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1e5+10;
typedef long long ll;
int n,dp[N][20],a[N];
struct hash_table {
	ll hash_mod = 590027;
	ll state[600000], ans[600000], up;
	ll tot, first[600000], nxt[600000], w[600000];
	void init() {
		memset(first, 0, sizeof(first));
		tot = 0;
		up = 0;
	}
	ll ins(ll sta, ll val) {
		ll key = sta%hash_mod;
		for(ll i = first[key]; i; i = nxt[i]) {
			if(state[w[i]] == sta) return ans[w[i]] += val;
		}
		state[++up] = sta;
		ans[up] = val;
		nxt[++tot] = first[key];
		w[tot] = up;
		first[key] = tot;
		return val;
	}
} mmp;
ll find(int key) {
	for(ll k = 1; k <= mmp.up; ++k) {
		ll sta = mmp.state[k];
		ll w = mmp.ans[k];
		if(sta==key)return w;
//        printf("sta:%lld w%lld\n",sta,w);
	}
	return 0;
}
ll gcd(ll a,ll b) {
	return (b==0)?a:gcd(b,a%b);
}
void ST() {
	for(int j=1; (1<<j)<=n; j++) {
		for(int i=1; i+(1<<j)-1<=n; i++) {
			dp[i][j]=gcd(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
		}
	}
}

int RMQ(int i,int j) {
	int k=0;
	while((1<<(k+1))<=j-i+1) k++;
	return gcd(dp[i][k],dp[j-(1<<k)+1][k]);
}

void solve() {
	for(int i=1; i<=n; i++) {
		int l=i,r=n,temp=a[i];
		while(l<=n) {
			int L=l;
			while(l<=r) {
				int mid=(r - l)/2 + l;
				if(RMQ(i,mid)==temp) {
					l=mid+1;
				} else {
					r=mid-1;
				}
			}
			//mmp[temp]+=(r-L+1);
			mmp.ins(temp,(r-L+1));
			temp=gcd(temp,a[l]);
			r=n;
		}
	}
}

int main() {
	int t;
	scanf("%d",&t);
	for(int T=1; T<=t; T++) {
		mmp.init();
		scanf("%d",&n);
		for(int i=1; i<=n; i++) {
			scanf("%d",&a[i]);
			dp[i][0]=a[i];
		}
		ST();
		solve();
		int q;
		scanf("%d",&q);
		printf("Case #%d:\n",T);
		while(q--) {
			int x,l,r;
			scanf("%d%d",&l,&r);
			x = RMQ(l,r);
			ll ans=find(x);
			printf("%d %lld\n",x,ans);
		}
	}
	return 0;
}
/*
1
5
1 2 4 6 7
4
1 5
2 4
3 4
4 4
*/