HDU 5634 Rikka with Phi (线段树)

Problem Description

Rikka and Yuta are interested in Phi function (which is known as Euler's totient function).

Yuta gives Rikka an array A[1..n] of positive integers, then Yuta makes m queries.  

There are three types of queries: 

1lr 

Change A[i] into φ(A[i]), for all i∈[l,r].

2lrx 

Change A[i] into x, for all i∈[l,r].

3lr 

Sum up A[i], for all i∈[l,r].

Help Rikka by computing the results of queries of type 3.

 

Input

The first line contains a number T(T≤100) ——The number of the testcases. And there are no more than 2 testcases with n>105

For each testcase, the first line contains two numbers n,m(n≤3×105,m≤3×105)。

The second line contains n numbers A[i]

Each of the next m lines contains the description of the query. 

It is guaranteed that 1≤A[i]≤107 At any moment.

 

Output

For each query of type 3, print one number which represents the answer.

 

Sample Input

1
10 10
56 90 33 70 91 69 41 22 77 45
1 3 9
1 1 10
3 3 8
2 5 6 74
1 1 8
3 1 9
1 2 10
1 4 9
2 8 8 69
3 3 9

 

Sample Output

80
122
86

在更新值变成欧拉函数相应的值的时候，效果最坏的情况是区间里每个点的值都不一样
那么区间的每个点都要遍历到。然后由于第2个操作会将一段区间的值都变成相同的，所以
在操作1的时候，可以不用更新每个点，只更新一下区间标记一下就可以。所以直接用线段树
暴力来搞，发现是不会超时的。另外线段树的区间要注意，n最大是3*1e5;
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long int LL;
const int maxn=1e5;
bool check[maxn*100+5];
LL prime[maxn*100+5];
LL eul[maxn*100+5];
LL sum[maxn*20+5];
LL a[maxn*20+5];
int n,m;
/*void eular(){
    eul[1] = 1;
    for (int i = 2; i<maxn*100+5; i++) eul[i] = i;
    for (int i = 2; i<maxn*100+5; i++)
        if (eul[i] == i)
            for (int j = i; j<maxn*100+5; j += i) eul[j] = eul[j] / i*(i - 1);
}
 */
void eular()
{
    memset(check,false,sizeof(check));
    eul[1]=1;
    int tot=0;
    for(int i=2;i<=maxn*100+5;i++)
    {
        if(!check[i])
        {
            prime[tot++]=i;
            eul[i]=i-1;
        }
        for(int j=0;j<tot;j++)
        {
            if(i*prime[j]>maxn*100+5) break;
            check[i*prime[j]]=true;
            if(i%prime[j]==0)
            {
                eul[i*prime[j]]=eul[i]*prime[j];
                break;
            }
            else
            {
                eul[i*prime[j]]=eul[i]*(prime[j]-1);
            }
            
        }
    }
}

void pushup(int node)
{
    sum[node]=sum[node<<1]+sum[node<<1|1];
    if(a[node<<1]==a[node<<1|1])
        a[node]=a[node<<1];
    else
        a[node]=0;
}
void pushdown(int node,int l,int r)
{
    int mid=(l+r)>>1;
    if(a[node])
    {
        sum[node<<1]=a[node]*(mid-l+1);
        sum[node<<1|1]=a[node]*(r-mid);
        a[node<<1]=a[node<<1|1]=a[node];
        //c[node<<1]=c[node<<1|1]=c[node];
        a[node]=0;
    }
}
void build(int node,int l,int r)
{
    if(l==r)
    {
        scanf("%lld",&a[node]);
        sum[node]=a[node];
        return ;
    }
    int mid=(l+r)>>1;
    build(node<<1,l,mid);
    build(node<<1|1,mid+1,r);
    pushup(node);
}
void update1(int node,int l,int r,int L,int R,LL tag)
{
    if(L<=l&&r<=R)
    {
        sum[node]=(LL)tag*(r-l+1);
        a[node]=tag;
        //c[node]=tag;
        return;
    }
    if(a[node]) pushdown(node,l,r);
    int mid=(l+r)>>1;
    if(L<=mid)
        update1(node<<1,l,mid,L,R,tag);
    if(R>mid)
        update1(node<<1|1,mid+1,r,L,R,tag);
    pushup(node);
}
void update2(int node,int l,int r,int L,int R)
{
    if(L<=l&&r<=R)
    {
        if(a[node])
        {
            a[node]=eul[a[node]];
            sum[node]=a[node]*(r-l+1);
            return;
        }
        
        int mid=(l+r)>>1;
        if(L<=mid)
            update2(node<<1,l,mid,L,R);
        if(R>mid)
            update2(node<<1|1,mid+1,r,L,R);
        pushup(node);
        return ;
    }
    if(a[node]) pushdown(node,l,r);
    int mid=(l+r)>>1;
    if(L<=mid)
        update2(node<<1,l,mid,L,R);
    if(R>mid)
        update2(node<<1|1,mid+1,r,L,R);
    pushup(node);
}
LL query(int node,int l,int r,int L,int R)
{
    if(L<=l&&r<=R)
    {
        return sum[node];
    }
    if(a[node]) pushdown(node,l,r);
    
    int mid=(l+r)>>1;
    LL ret=0;
    if(L<=mid)
        ret+=query(node<<1,l,mid,L,R);
    if(R>mid)
        ret+=query(node<<1|1,mid+1,r,L,R);
   
    return ret;
}
int main()
{
    int t;
    scanf("%d",&t);
    int x,y,z;
    LL w;
    eular();
    while(t--)
    {
        scanf("%d%d",&n,&m);
        build(1,1,n);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            if(x==1)
            {
                update2(1,1,n,y,z);
            }
            else if(x==2)
            {
                scanf("%lld",&w);
                update1(1,1,n,y,z,w);
            }
            else
            {
                printf("%lld
",query(1,1,n,y,z));
            }
        }
    }
    return 0;
}