Ralph is in the Binary Country. The Binary Country consists of n cities and (n - 1) bidirectional roads connecting the cities. The roads are numbered from 1 to (n - 1), the i-th road connects the city labeled 
(here ⌊ x⌋ denotes the x rounded down to the nearest integer) and the city labeled (i + 1), and the length of the i-th road is Li.
Now Ralph gives you m queries. In each query he tells you some city Ai and an integer Hi. He wants to make some tours starting from this city. He can choose any city in the Binary Country (including Ai) as the terminal city for a tour. He gains happiness (Hi - L) during a tour, where L is the distance between the city Ai and the terminal city.
Ralph is interested in tours from Ai in which he can gain positive happiness. For each query, compute the sum of happiness gains for all such tours.
Ralph will never take the same tour twice or more (in one query), he will never pass the same city twice or more in one tour.
The first line contains two integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 105).
(n - 1) lines follow, each line contains one integer Li (1 ≤ Li ≤ 105), which denotes the length of the i-th road.
m lines follow, each line contains two integers Ai and Hi (1 ≤ Ai ≤ n, 0 ≤ Hi ≤ 107).
Print m lines, on the i-th line print one integer — the answer for the i-th query.
2 2
5
1 8
2 4
11
4
6 4
2
1
1
3
2
2 4
1 3
3 2
1 7
11
6
3
28
Here is the explanation for the second sample.
Ralph's first query is to start tours from city 2 and Hi equals to 4. Here are the options:
- He can choose city 5 as his terminal city. Since the distance between city 5 and city 2 is 3, he can gain happiness 4 - 3 = 1.
- He can choose city 4 as his terminal city and gain happiness 3.
- He can choose city 1 as his terminal city and gain happiness 2.
- He can choose city 3 as his terminal city and gain happiness 1.
- Note that Ralph can choose city 2 as his terminal city and gain happiness 4.
- Ralph won't choose city 6 as his terminal city because the distance between city 6 and city 2 is 5, which leads to negative happiness for Ralph.
So the answer for the first query is 1 + 3 + 2 + 1 + 4 = 11.
大致题意：m次询问，每次给定一个a和h,第i个点的贡献是h减i到a的距离，求大于0的贡献.
分析：非常棒的一道题!改变了我对cf测评机的认识.我一开始错误地认为只需要求出一个点祖先的贡献和它的子树的节点的贡献就好了，没有考虑到还有一种情况：它的父亲的其它儿子的子树可能还有贡献.这样的话就必须求出每个点为根的子树中每个点到根节点的距离.事实上只需要存下距离就够了，因为我们不关心到底是哪一个点.我原本以为内存会爆掉，因为n有10^6,要把每个点都给保存下来，还有保存子树的所有点的距离，数组绝对是开不下的，但是如果不求出这个是无法继续做下去的，于是用vector，竟然没有爆内存，神奇.
因为题目给定的是一个二叉树，每个节点的编号都是有规律的，所以可以从第n号点从下往上更新，利用左右儿子的信息去更新根节点的信息.为了在查询的时候方便一些，可以维护一下前缀和，即第i个点的子树中前j个距离的和，这样可以利用vector的upper_bound来快速查找关键点p,利用公式进行O(1)计算.
接下来的事情就很好办了，每次从查询的点开始，先看看左右子树之前有没有被查询，如果没有就查询一下，完了之后就跳到祖先，并记录上一次查询的是哪一个点，下次就不查询它了.
参考了网上神犇的vector的写法，用得很6，个人认为它最大的用处是关于内存方面的.这道题也纠正了我的一个常识性错误：if (k) ......我一直以为如果k > 0这个判断语句才会成立，万万没想到是k != 0这个语句就成立了，看来还是基本功不扎实,要多多练习.
#include <cstdio> #include <vector> #include <cstring> #include <iostream> #include <algorithm>using namespace std;typedef long long ll;const int maxn = 1000010; vector <ll>e[maxn], sum[maxn]; ll n, m, len[maxn], a, h; ll ans;ll query(ll x, ll h) {if (h <= 0)return 0;ll p = upper_bound(e[x].begin(), e[x].end(), h) - e[x].begin();return p * h - sum[x][p - 1]; }void init() {for (int i = n; i >= 1; i--){e[i].push_back(0);int lc = i * 2, rc = i * 2 + 1;if (lc <= n){for (int j = 0; j < e[lc].size(); j++)e[i].push_back(e[lc][j] + len[lc]);}if (rc <= n){for (int j = 0; j < e[rc].size(); j++)e[i].push_back(e[rc][j] + len[rc]);}sort(e[i].begin(), e[i].end());sum[i].resize(e[i].size());for (int j = 1; j < sum[i].size(); j++)sum[i][j] = sum[i][j - 1] + e[i][j];} }int main() {cin >> n >> m;for (int i = 2; i <= n; i++)scanf("%I64d", &len[i]);init();while (m--){ans = 0;scanf("%I64d%I64d", &a, &h);ll last = 0;while (a && h >= 1){ans += h;ll lc = a * 2, rc = a * 2 + 1;if (last != lc && lc <= n)ans += query(lc, h - len[lc]);if (last != rc && rc <= n)ans += query(rc, h - len[rc]);last = a;h -= len[a];a /= 2;}printf("%I64d\n", ans);}return 0; }
