P5142区间方差求调,未过样例
查看原帖
P5142区间方差求调,未过样例
607952
ZHANGGUIZHI楼主2024/10/16 19:41
#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 1e5 + 3;
const int P = 1e9 + 7;
int n, m, a[N], Mod;
int qinv(int a, int p, int mod) {
    int x = 1;
    for (; p; p >>= 1) {
        if (p & 1) x = 1 ll * x * a % mod;
        a = 1 ll * a * a % mod;
    }
    return x;
}
struct node {
    int l, r, vals, valp;
}
t[N << 2];
void pushup(int p) {
    t[p].vals = (t[p << 1].vals + t[p << 1 | 1].vals) % P;
    t[p].valp = (t[p << 1].valp + t[p << 1 | 1].valp) % P;
}
void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    if (l == r) {
        t[p].vals = a[l];
        t[p].valp = 1 ll * a[l] * a[l] % P;
        return;
    }
    int mid = t[p].l + t[p].r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}
void change(int p, int x, int k) {
    if (t[p].l == t[p].r && t[p].l == x) {
        t[p].vals = k;
        t[p].valp = 1 ll * k * k % P;
        return;
    }
    int mid = t[p].l + t[p].r >> 1;
    if (x <= mid) change(p << 1, x, k);
    if (x > mid) change(p << 1 | 1, x, k);
    pushup(p);
}
int askfc(int p, int L, int R) {
    int ans = 0;
    if (t[p].l >= L && t[p].r <= R) {
        return t[p].valp;
    }
    int mid = t[p].l + t[p].r >> 1;
    if (L <= mid) ans = (ans + askfc(p << 1, L, R)) % P;
    if (R > mid) ans = (ans + askfc(p << 1 | 1, L, R)) % P;
    return ans;
}
int asksum(int p, int L, int R) {
    int ans = 0;
    if (t[p].l >= L && t[p].r <= R) {
        return t[p].vals;
    }
    int mid = t[p].l + t[p].r >> 1;
    if (L <= mid) ans = (ans + asksum(p << 1, L, R)) % P;
    if (R > mid) ans = (ans + asksum(p << 1 | 1, L, R)) % P;
    return ans;
}
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    build(1, 1, n);
    int op, x, y;
    while (m--) {
        cin >> op >> x >> y;
        if (op == 1) change(1, x, y);
        else {
            Mod = qinv(y - x + 1, P - 2, P);
            int fc = askfc(1, x, y);
            int sum = asksum(1, x, y);
            cout << (1 ll * fc * Mod % P) - (1 ll * sum * Mod % P) << '\n';
        }
    }
    return 0;
}
2024/10/16 19:41
加载中...