https://nanti.jisuanke.com/t/42387
The meaning of problems: n (1 <= n <= 1e5) number, initially 1 with q (1 <= q <= 1e5) query times, both. 1, mul l, r, x interval [l, r] is multiplied by x (1 <= x <= 10). 2, query l, r interrogation interval [l, r] each prime factor factorization maximum value of the maximum index.
Solution: After understanding the problem know, is in the prime number of statistics for each interval (2, 3, 5, 7) number, the maximum range of the maximum value of the index number for each prime number.
Maintenance of four tree line (2,3,5,7) statistics are the number of prime numbers, the maximum range of queries every tree, in the maximum four trees is the answer.
#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define cin(x) scanf("%lld" , &x);
using namespace std;
const int N = 1e7+9;
const int maxn = 1e5+9;
const double esp = 1e-2;
int q , n;
char s[20];
struct node{
int l , r , ma, lazy;
};
struct Segment_Tree{
node tree[maxn<<2];
void pushup(int root){
tree[root].val = tree[root>>1].val + tree[root>>1|1].val;
tree[root].ma = max(tree[root<<1].ma , tree[root<<1|1].ma);
}
void pushdown(int root){
tree[root<<1].ma += tree[root].lazy;
tree[root<<1|1].ma += tree[root].lazy;
tree[root<<1].lazy += tree[root].lazy;
tree[root<<1|1].lazy += tree[root].lazy;
tree[root].lazy = 0;
}
void build(int l , int r , int root){
tree[root].l = l , tree[root].r = r , tree[root].val = 0, tree[root].lazy = 0 , tree[root].ma = 0;
if(l == r){
return ;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
}
void update(int l , int r , int root , int va){
if(tree[root].l >= l && tree[root].r <= r){
tree[root].lazy += va ;
tree[root].ma += va;
return ;
}
if(tree[root].lazy) pushdown(root);
int mid = (tree[root].l + tree[root].r) >> 1;
if(l <= mid)
update(l , r , root<<1 , va);
if(r > mid)
update(l , r , root<<1|1 , va);
pushup(root);
}
int query(int l , int r , int root){
int ma = -INF;
if(tree[root].l >= l && tree[root].r <= r){
return tree[root].ma;
}
if(tree[root].lazy) pushdown(root);
int mid = (tree[root].l + tree[root].r) >> 1;
if(mid >= l)
ma = max(ma , query(l , r , root<<1));
if(r > mid)
ma = max(ma , query(l , r , root<<1|1));
return ma ;
}
}pr2 , pr3 , pr5 , pr7;
void solve(){
scanf("%lld%lld" , &n , &q);
pr2.build(1 , n , 1);
pr3.build(1 , n , 1);
pr5.build(1 , n , 1);
pr7.build(1 , n , 1);
rep(i , 1 , q){
scanf("%s" , s);
if(s[0] == 'M' && s[1] == 'U'){
int l , r , x;
scanf("%lld%lld%lld" , &l , &r , &x);
if(x == 2){
pr2.update(l , r , 1,1);
}else if(x == 3){
pr3.update(l , r , 1,1);
}else if(x == 4){
pr2.update(l , r , 1,2);
}else if(x == 5){
pr5.update(l , r , 1,1);
}else if(x == 6){
pr2.update(l , r , 1,1);
pr3.update(l , r , 1,1);
}else if(x == 7){
pr7.update(l , r , 1,1);
}else if(x == 8){
pr2.update(l , r , 1,3);
}else if(x == 9){
pr3.update(l , r , 1,2);
}else if(x == 10){
pr2.update(l , r , 1,1);
pr5.update(l , r , 1,1);
}
}else{
int l , r ;
int ans = -1 ;
scanf("%lld%lld" , &l , &r);
ans = max(ans , pr2.query(l , r , 1));
ans = max(ans , pr3.query(l , r , 1));
ans = max(ans , pr5.query(l , r , 1));
ans = max(ans , pr7.query(l , r , 1));
printf("ANSWER %lld\n" , ans);
}
}
}
signed main()
{
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
//int t ;
//cin(t);
//while(t--){
//while(~scanf("%lld%lf" , &n , &a))
solve();
//}
}