[Bzoj5358][Lydsy1805月赛]口算训练（预处理+动态开点线段树）

5358: [Lydsy1805月赛]口算训练

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Time Limit: 5 Sec  Memory Limit: 512 MB
Submit: 318  Solved: 105
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Description

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begin.lydsy.com/JudgeOnline/upload/201805.pdf

Input

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Output

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Sample Input

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Sample Output

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HINT

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Source

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鸣谢claris提供

分析：

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首先看到数最大只有10w，美滋滋，筛出素数，预处理出每个数包含哪些质数，且包含多少个。

然后对于每个质数开棵线段树。

d = （p1^c1）（p2^c2）……（pk^ck），

查询的时候只用查询d的每个质数在[L,R]范围内出现次数是否大于d中次数就行

AC代码：

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# include <iostream>
# include <cstdio>
# include <cstring>
# include <vector>
# include <algorithm>
using namespace std;
const int N = 1e5;
vector<pair<int,int> > t[N + ];
int prime[N + ],cnt,root[N + ],lc[N * ],rc[N * ],s[N * ],p,a[N + ];bool vis[N + ];
void shai()
{
    for(int i = ;i <= N;i++)
    {
        if(!vis[i])prime[++cnt] = i;
        for(int j = ;j <= cnt;j++)
        {
            if(i * prime[j] > N)break;
            vis[i * prime[j]] = true;
            if(i % prime[j] == )break;
        }
    }
    for(int i = ;i <= cnt;i++)
     for(int j = prime[i];j <= N;j += prime[i])
     {
        t[j].push_back(make_pair(prime[i],j / prime[i]));
     }
}
void updata(int &k,int L,int l,int r,int d)
{
    if(!k){k = ++p;lc[k] = rc[k] = s[k] = ;}
    s[k] += d;
    if(l == r)return;
    int mid = l + r >> ;
    if(L <= mid)updata(lc[k],L,l,mid,d);
    else updata(rc[k],L,mid + ,r,d);
}
int Query(int &k,int L,int R,int l,int r)
{
    if(!k)return ;
    if(L <= l && r <= R)return s[k];
    int mid = l + r >> ;
    if(L > mid)return Query(rc[k],L,R,mid + ,r);
    if(R <= mid)return Query(lc[k],L,R,l,mid);
    return Query(lc[k],L,R,l,mid) + Query(rc[k],L,R,mid + ,r);
}
int n,m;
int main()
{
 shai();int Case;
 scanf("%d",&Case);
 while(Case--)
 {
    memset(root,,sizeof root);p = ;
    scanf("%d %d",&n,&m);
    for(int i = ;i <= n;i++)
    {
        scanf("%d",&a[i]);
        for(int j = ;j < (int)t[a[i]].size();j++)
        updata(root[t[a[i]][j].first],i,,n,t[a[i]][j].second);
    }
    int x,y,z;
    while(m--)
    {
        scanf("%d %d %d",&x,&y,&z);bool f = true;
        for(int j = ;j < (int)t[z].size();j++)
        if(Query(root[t[z][j].first],x,y,,n) < t[z][j].second){f = false;break;}
        if(f)puts("Yes");else puts("No");
    }
 }
}