花神游历各国

题目描述

思路

1. 0 和 1 开方之后不会改变，对 0， 1 节点打上标记
2. 如果孩子节点都打上标记了，说明这些孩子节点的和不会改变，那么父节点也打上标记
3. 对于打上标记的点，change 的时候直接 return

代码

#include <cstdio>
#include <cmath>

int n, m;
long long arr[100005 << 2], at[100005];
bool flag[100005 << 2];
inline int read() {
    int s = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s;
}
void build(int k, int l, int r) {
    if (l == r) {
        arr[k] = at[l];
        if (arr[k] <= 1) flag[k] = 1;
        return;
    }
    int mid = l + r >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    arr[k] = arr[k << 1] + arr[k << 1 | 1];
    if (flag[k << 1] && flag[k << 1| 1]) flag[k] = 1;
}
void change(int k, int l, int r, int x, int y) {
    if (flag[k]) return;
    if (l == r) {
        arr[k] = sqrt(arr[k]);
        if (arr[k] <= 1) flag[k] = 1;
        return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid) change(k << 1, l, mid, x, y);
    if (y > mid) change(k << 1 | 1, mid + 1, r, x, y);
    arr[k] = arr[k << 1] + arr[k << 1 | 1];
    if (flag[k << 1] && flag[k << 1| 1]) flag[k] = 1;
}
long long query(int k, int l, int r, int x, int y) {
    if (x <= l && r <= y) return arr[k];
    int mid = (l + r) >> 1;
    long long res = 0;
    if (x <= mid) res = query(k << 1, l, mid, x, y);
    if (y > mid) res += query(k << 1 | 1, mid + 1, r, x, y);
    return res;
}
int main() {
    n = read();
    for (int i = 1, j; i <= n; ++i) at[i] = read();
    m = read();
    build(1, 1, n);
    for (int i = 1, a, b, c; i <= m; ++i) {
        a = read(), b = read(), c = read();
        if (a == 1) {
            printf("%lld\n", query(1, 1, n, b, c));
        } else {
            change(1, 1, n, b, c);
        }
    }
    return 0;
}

花神游历各国

原文：https://www.cnblogs.com/liuzz-20180701/p/11495917.html