Rikka with Phi

 Accepts: 5
 Submissions: 66
 Time Limit: 16000/8000 MS (Java/Others)
 Memory Limit: 131072/131072 K (Java/Others)
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]A[1..n] of positive integers, then Yuta makes mm queries.

There are three types of queries:

1 \; l \; r1lr

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

2 \; l \; r \; x2lrx

Change A[i]A[i] into xx, for all i \in [l, r]i∈[l,r].

3 \; l \; r3lr

Sum up A[i]A[i], for all i \in [l, r]i∈[l,r].

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

Input

The first line contains a number T(T \leq 100)T(T≤100) ——The number of the testcases. And there are no more than 2 testcases with n > 10 ^ 5n>10​5​​

For each testcase, the first line contains two numbers n,m(n \leq 3 \times 10^5, m \leq 3 \times 10^5)n,m(n≤3×10​5​​,m≤3×10​5​​)。

The second line contains nn numbers A[i]A[i]

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

It is guaranteed that 1 \leq A[i] \leq 10^71≤A[i]≤10​7​​ 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
/*
BestCoder Round #73 (div.1)
hdu5634 Rikka with Phi
本来最开始用伸展树(主要是线段树不是很熟)的,但是中间有点问题,导致一直是TLE,于是乎
就去写线段树了
感觉对线段树的理解上有很多问题TAT
思路:先打个表然后用线段树(平衡树)去解决
主要是有same标记,sum和,欧拉
中间在将一个区间改变成欧拉时,如果中间遇到了区间的same,直接
修改这个标记就好的
hhh-2016-02-25 17:54:59
*/ #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
using namespace std;
typedef long long ll;
typedef long double ld; const int maxn = 300010;
ll euler[10000010] = {0};
void getEuler()
{
euler[1] = 1;
for(ll i =2 ; i <= 10000000; i++)
{
if(!euler[i])
{
for(ll j = i; j <= 10000000; j += i)
{
if(!euler[j])
euler[j] = j;
euler[j] = euler[j]/i*(i-1);
}
}
}
} struct node
{
int l,r;
ll sum,same;
} segtree[maxn<<2]; void push_up(int r)
{
int lson = r<<1,rson = (r<<1)|1;
segtree[r].sum = segtree[lson].sum + segtree[rson].sum;
if(segtree[lson].same == segtree[rson].same)
segtree[r].same = segtree[lson].same;
else
segtree[r].same = 0;
} void build(int i ,int l,int r)
{
segtree[i].l = l,segtree[i].r = r;
segtree[i].sum = segtree[i].same = 0;
if(l == r)
{
scanf("%I64d",&segtree[i].same);
segtree[i].sum = segtree[i].same;
return ;
}
int mid = (l+r)>>1;
build(i<<1,l,mid);
build((i<<1)|1,mid+1,r);
push_up(i);
} void push_down(int r)
{
int lson = r<<1,rson = (r<<1)|1;
if(segtree[r].same)
{
segtree[lson].same = segtree[r].same;
segtree[rson].same = segtree[r].same;
segtree[lson].sum = (ll)(segtree[lson].r - segtree[lson].l+1)*segtree[r].same;
segtree[rson].sum = (ll)(segtree[rson].r - segtree[rson].l+1)*segtree[r].same;
segtree[r].same = 0;
}
} void make_euler(int i,int l,int r)
{
//区间修改
if(segtree[i].r <= r && segtree[i].l >= l && segtree[i].same)
{
segtree[i].same = (ll)euler[segtree[i].same];
segtree[i].sum = segtree[i].same*(ll)(segtree[i].r-segtree[i].l+1);
return;
}
if(l == r) return ;
int mid = (segtree[i].r +segtree[i].l) >>1;
push_down(i);
if(l <= mid)
make_euler(i<<1,l,r);
if(r > mid)
make_euler((i<<1)|1,l,r);
push_up(i);
}
void make_same(int i,int l,int r,ll c)
{
if(segtree[i].r <= r && segtree[i].l >= l)
{
segtree[i].same = c;
segtree[i].sum = segtree[i].same*(ll)(segtree[i].r-segtree[i].l+1);
return;
}
int mid = (segtree[i].r +segtree[i].l) >>1;
push_down(i);
if(l <= mid)
make_same(i<<1,l,r,c);
if(r > mid)
make_same((i<<1)|1,l,r,c);
push_up(i);
} ll get_sum(int i,int l,int r)
{
int mid = (segtree[i].l + segtree[i].r) >> 1;
if(segtree[i].l >= l && segtree[i].r <= r)
{
return segtree[i].sum;
}
push_down(i);
ll ans = 0;
if(l <= mid) ans += get_sum(i<<1,l,r);
if(r > mid) ans += get_sum(i<<1|1,l,r);
push_up(i);
return ans;
} int main()
{
int T,n,m;
getEuler();
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
build(1,1,n);
for(int i =1; i <= m; i++)
{
int op,l,r;
ll c;
scanf("%d",&op);
if(op == 1)
{
scanf("%d%d",&l,&r);
make_euler(1,l,r);
//debug();
}
else if(op == 2)
{
scanf("%d%d%I64d",&l,&r,&c);
make_same(1,l,r,c);
// debug();
}
else if(op == 3)
{
scanf("%d%d",&l,&r);
printf("%I64d\n",get_sum(1,l,r));
}
}
}
return 0;
}

  

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