数列分块入门 5

分块训练

参考hzwer的做法，每个数在经过数次向下取整的开方后都会变成0或1。

我们在维护块内信息的时候，可以将整块全是0或1的块跳过，减少复杂度。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 50005;
ll a[N], sum[N];
int n, t, lt[N], rt[N], pos[N];
bool vis[N];

void reset(int k){
    if(vis[k]) return;
    vis[k] = true, sum[k] = 0;
    for(int i = lt[k]; i <= rt[k]; i ++){
        a[i] =(int)sqrt(a[i]);
        sum[k] += a[i];
        if(a[i] > 1) vis[k] = false;
    }
}

void modify(int l, int r){
    int p = pos[l], q = pos[r];
    if(p == q){
        for(int i = l; i <= r; i ++){
            sum[p] -= a[i], a[i] = (int)sqrt(a[i]), sum[p] += a[i];
        }
    }
    else{
        for(int i = p + 1; i <= q - 1; i ++)
            reset(i);
        for(int i = l; i <= rt[p]; i ++){
            sum[p] -= a[i], a[i] = (int)sqrt(a[i]), sum[p] += a[i];
        }
        for(int i = lt[q]; i <= r; i ++){
            sum[q] -= a[i], a[i] = (int)sqrt(a[i]), sum[q] += a[i];
        }
    }
}

ll query(int l, int r){
    int p = pos[l], q = pos[r];
    ll ret = 0;
    if(p == q){
        for(int i = l; i <= r; i ++){
            ret += a[i];
        }
    }
    else{
        for(int i = p + 1; i <= q - 1; i ++) ret += sum[i];
        for(int i = l; i <= rt[p]; i ++) ret += a[i];
        for(int i = lt[q]; i <= r; i ++) ret += a[i];
    }
    return ret;
}

int main(){

    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    cin >> n;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    t = (int)sqrt(n);
    for(int i = 1; i <= t; i ++){
        lt[i] = (i - 1) * t + 1;
        rt[i] = i * t;
    }
    if(rt[t] < n) t ++, lt[t] = rt[t - 1] + 1, rt[t] = n;
    for(int i = 1; i <= t; i ++){
        for(int j = lt[i]; j <= rt[i]; j ++){
            pos[j] = i, sum[i] += a[j];
        }
    }
    for(int i = 1; i <= n; i ++){
        int opt, l, r; ll c;
        cin >> opt >> l >> r >> c;
        if(!opt) modify(l, r);
        else cout << query(l, r) << endl;
    }
    return 0;
}