rt,本题我用线段树在洛谷跑的是一个点60ms一下(AC),然后在机构OJ(数据范围询问次数乘了2)提交一大堆TLE(都能跑到2000ms朝上),卡了卡常发现并没有什么作用,请问这是咋回事?是我有啥没考虑到吗,,,
code
#include<bits/stdc++.h>
#define reg register int
#define INF (1<<30)
#define pb push_back
#define vc vector
#define fst first
#define scd second
#define int long long
#define rep(i,x,y) for(int i=x;i<=y;i++)
using namespace std;
int read(){
int res=0,fs=1; char c=getchar();
while(!(c>='0' && c<='9')){ if(c=='-')fs=-1; c=getchar(); }
while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
return res*fs;
}
void print(int x){
if(x<0) { putchar('-'); x=-x;}
if(x>9) print(x/10);
putchar(x%10+'0');
}
int n,cnt,m,a[500010],ans,tmp,k;
typedef pair<int,int> P;
//struct PROBLEM_SOLVER{
// int n,m;
//}solver;
//signed main(){
void py(){
cout<<"YES\n";
}
void pn(){
cout<<"NO\n";
}
const int maxn=1e5+10;
struct tree{
int s[4*maxn],lazy[4*maxn];
void build(int p,int l,int r){
if(l==r){
s[p]=a[l];
return ;
}
int m=(l+r)>>1;
build(p<<1,l,m);
build((p<<1)+1,m+1,r);
s[p]=s[(p<<1)]+s[(p<<1)+1];
}
void update(int p,int l,int r,int x,int y){
if(r<x||l>y) return ;
if(lazy[p]==1) return ;
if(x<=l&&r<=y){
int fl=0;
for(int i=l;i<=r;++i){
a[i]=sqrt(a[i]);
if(a[i]!=1) fl=1;
}
if(fl==0) lazy[p]=1;
build(p,l,r);
return ;
}
int m=(l+r)>>1;
update(p<<1,l,m,x,y);
update((p<<1)+1,m+1,r,x,y);
s[p]=s[p<<1]+s[1+(p<<1)];
}
int query(int p,int l,int r,int x,int y){
if(r<x||l>y) return 0;
if(x<=l&&r<=y){
return s[p];
}
int m=(l+r)>>1;
return query(p<<1,l,m,x,y)+query((p<<1)+1,m+1,r,x,y);
}
}t;
signed main() {
cin>>n;
for(int i=1;i<=n;i++) a[i]=read();
t.build(1,1,n);
cin>>m;
while(m--){
int op,l,r;
// cin>>op>>l>>r;
op=read(),l=read(),r=read();
if(l>r) swap(l,r);
if(op==1){
cout<<t.query(1,1,n,l,r)<<'\n';
}else{
t.update(1,1,n,l,r);
}
}
return 0;
}