codeforces 920F

題目描述

令\(d(x)\)為正整數\(x\)的因子的數量。給定一個長度為\(n\)的序列\(a\)，有兩種操作：

1. 將\([l,r]\)的數i變為\(d(i)\)
2. 求\(\sum_{i=l}^{r}{a_i}\)

思路

首先可以用線段樹來維護這個序列，每次暴力修改。因為\(d(1)=1,d(2)=2\)，因此當一個區間的最大值小於2時就不需要修改。又因為每次修改只會讓值減少，而每個數最多會被修改\(O(log_2n)\)次，因此能夠保證復雜度

代碼

#include <bits/stdc++.h>

using namespace std;

typedef
 
 long long ll;

const int MAXN = 300000 + 10;
const int MAXA = 1000000 + 10;
const char ENDLINE = ‘\n‘;

int d[MAXA], num[MAXA];
int prime[MAXA / 10], primeCnt;

void sieve()
{
    const int MX = 1000000;
    static bool vis[MAXA];
    d[1] = 1;
    for (int i = 2; i <= MX; ++i) {
        if (!vis[i]) {
            prime[++primeCnt] = i;
            d[i] = 2
 
;
            num[i] = 1;
        }
        for (int j = 1; j <= primeCnt && prime[j] * i <= MX; ++j) {
            vis[prime[j] * i] = true;
            if (i % prime[j] == 0) {
                num[i * prime[j]] = num[i] + 1;
                d[i * prime[j]] = d[i] / (num[i] + 1) * (num[i] + 2
 
);
                break;
            }
            num[i * prime[j]] = 1;
            d[i * prime[j]] = d[i] * 2;
        }
    }

}

struct SegmentTree {
    int L[MAXN << 2], R[MAXN << 2];
    int mx[MAXN << 2];
    ll sum[MAXN << 2];
    void pushup(int o) { mx[o] = max(mx[o<<1], mx[o<<1|1]), sum[o] = sum[o<<1] + sum[o<<1|1]; }
    void update(int l, int r, int o=1)
    {
        if (mx[o] <= 2) 
            return;
        if (L[o] == R[o]) {
            sum[o] = mx[o] = d[sum[o]];
            return;
        }
        int mid = (L[o] + R[o]) >> 1;
        if (l <= mid)
            update(l, r, o<<1);
        if (mid < r)
            update(l, r, o<<1|1);
        pushup(o);
    }
    ll query(int l, int r, int o=1)
    {
        if (L[o] == l && R[o] == r)
            return sum[o];
        int mid = (L[o] + R[o]) >> 1;
        if (r <= mid)
            return query(l, r, o<<1);
        if (mid < l)
            return query(l, r, o<<1|1);
        return query(l, mid, o<<1) + query(mid+1, r, o<<1|1);
    }
    void build(int l, int r, int a[], int o=1)
    {
        L[o] = l, R[o] = r;
        if (l == r) {
            sum[o] = mx[o] = a[l];
            return;
        }
        int mid = (l + r) >> 1;
        build(l, mid, a, o<<1);
        build(mid+1, r, a, o<<1|1);
        pushup(o);
    }
} segt;

int main()
{
    int n, m;
    static int a[MAXN];
    ios::sync_with_stdio(0);
    sieve();
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];
    segt.build(1, n, a);
    for (int i = 1; i <= m; ++i) {
        int opt, l, r;
        cin >> opt >> l >> r;
        if (opt == 1) {
            segt.update(l, r);
        } else {
            cout << segt.query(l, r) << ENDLINE;
        }
    }
}

codeforces 920F SUM and REPLACE