题意
给出一个长为 nn 的数列 ,以及 nn 个操作,操作涉及区间开方,区间求和。
题解
维护区间和。对于区间开方操作,维护区间最小值,若区间最小值小于等于1,那么就不需要进行操作。否则暴力更新区间最小值和区间和。
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int nmax = 5e4+100;
const int INF = 0x3f3f3f3f;
const int mm = sqrt(5e4+100)+10;
int n,m;
ll data[nmax],mmax[mm],sum[mm];
int L[mm],R[mm],belong[nmax],num,block;
void init(){
block = 2*sqrt(n);
num = n / block; if(n % block) num++;
for(int i = 1;i<=num;++i)
L[i] = (i-1)*block+1,
R[i] = i*block,
mmax[i] = -INF;
R[num] = n;
}
int main() {
scanf("%d",&n);
init();
for(int i = 1;i<=n;++i){
scanf("%lld",&data[i]);
belong[i] = (i-1) / block + 1;
mmax[belong[i]] = max(mmax[belong[i]],data[i]);
sum[belong[i]] += data[i];
}
int l,r,op; ll w;
for(int i = 1;i<=n;++i){
scanf("%d %d %d %lld",&op,&l,&r,&w);
if(op == 0){
if(mmax[belong[l]] > 1){
for(int j = l;j<=min(r,R[belong[l]]);++j) data[j] = sqrt(data[j]);
mmax[belong[l]] = -INF, sum[belong[l]] = 0;
for(int j = L[belong[l]];j<=R[belong[l]];++j)
mmax[belong[l]] = max(mmax[belong[l]],data[j]),
sum[belong[l]] += data[j];
}
if(belong[r] != belong[l] && mmax[belong[r]] > 1){
for(int j = L[belong[r]];j<=r;++j) data[j] = sqrt(data[j]);
mmax[belong[r]] = -INF, sum[belong[r]] = 0;
for(int j = L[belong[r]];j<=R[belong[r]];++j){
mmax[belong[r]] = max(mmax[belong[r]],data[j]);
sum[belong[r]] += data[j];
}
}
for(int j = belong[l] + 1;j<=belong[r]-1;++j){
if(mmax[j] > 1){
mmax[j] = -INF, sum[j] = 0;
for(int k = L[j];k<=R[j];++k){
data[k] = sqrt(data[k]);
mmax[j] = max(mmax[j],data[k]);
sum[j] += data[k];
}
}
}
}else{
ll ans = 0;
for(int j = l;j<=min(r,R[belong[l]]);++j) ans += data[j];
if(belong[r] != belong[l] )for(int j = L[belong[r]];j<=r;++j) ans += data[j];
for(int j = belong[l]+1;j<=belong[r]-1;++j) ans += sum[j];
printf("%lld\n",ans);
}
}
return 0;
}