hdu 6287 口算训练（可持久化线段树+线性筛）

Posted 2021-09-17 jpphy0

目录

• 问题
• 分析
• 代码

问题

hdu 6287 口算训练 - https://acm.hdu.edu.cn/showproblem.php?pid=6287

分析

• 通过素因子的数量判断是否整除
• 使用主席树维护区间素因子数量
• 线性筛维护素因子
• 线性筛维护素因子的序
• 线性筛分解素因子

代码

#include<bits/stdc++.h>
using namespace std;
const int MXN = 1e5+5;
int N, M, tot = -1, ver[MXN];
int pri[MXN], vis[MXN], pos[MXN], n = 0; // 素数表、最大素因子及其序号
struct TreeNode{
	int l, r; // 左右子树
	int s; // 素因子出现次数
}tree[MXN<<8];
void init(){
	memset(pri, 0, sizeof pri);
	memset(vis, 0, sizeof vis);
	for(int i = 2; i < MXN; ++i){
		if(!vis[i]) pri[++n] = i, vis[i] = i, pos[i] = n;
		for(int j = 1; j <= n; ++j){
			if(i*pri[j] >= MXN || pri[j] > vis[i]) break;
			vis[i*pri[j]] = pri[j], pos[i*pri[j]] = j;
		}
	}
}
int build(int l, int r){
	int root = ++tot;
	TreeNode &rt = tree[root];
	rt.s = 0, rt.l = root, rt.r = root;
	return root;
}
int update(int node, int l, int r, int p, int val){
	int root = ++tot;
	TreeNode &rt = tree[root];
	rt = tree[node];
	if(l == r){ 
		rt.s += val;
		return root;
	}
	int mid = (l+r)>>1;
	if(p <= mid) rt.l = update(rt.l, l, mid, p, val);
	else rt.r = update(rt.r, mid+1, r, p, val);
	rt.s = tree[rt.l].s + tree[rt.r].s;
	return root;
}
int query(int last, int now, int l, int r, int p){
	TreeNode &t1 = tree[now], &t2 = tree[last];
	if(l == r)
		return t1.s - t2.s;	
	int mid = (l+r)>>1;
	if(p <= mid) return query(t2.l, t1.l, l, mid, p);
	else return query(t2.r, t1.r, mid+1, r, p);
}
int main(){
	int t, x, y, z, p, l, r, ok;
	init();
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &N, &M);
		memset(ver, 0, sizeof ver);
		ver[0] = build(1, n);
		for(int i = 1; i <= N; ++i){
			ver[i] = ver[i-1];
			scanf("%d", &x);
			while(x > 1){
				y = 0, z = vis[x], p = pos[x];
				while(x%z == 0) ++y, x = x/z;
				ver[i] = update(ver[i], 1, n, p, y);
			}
		}
		for(int i = 1; i <= M; ++i){
			scanf("%d%d%d", &l, &r, &x);
			ok = 1;
			while(x > 1){
				y = 0, z = vis[x], p = pos[x];
				while(x%z == 0) ++y, x = x/z;
				if(y > query(ver[l-1], ver[r], 1, n, p)) ok = 0, x = 1;
			}
			printf("%s\\n", ok?"Yes":"No");
		}
	}
    return 0;
}