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  • 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
    ", query(1, 1, n, b, c));
    		} else {
    			change(1, 1, n, b, c);
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/liuzz-20180701/p/11495917.html
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