2021“MINIEYE杯”中国大学生算法设计超级联赛（3）1011. Segment Tree with Pruning（记忆化搜索）

Problem Description

Chenjb is struggling with data stucture now. He is trying to solve a problem using segment tree. Chenjb is a freshman in programming contest, and he wrote down the following C/C++ code and ran ''𝙽𝚘𝚍𝚎* 𝚛𝚘𝚘𝚝 = 𝚋𝚞𝚒𝚕𝚍(𝟷, 𝚗)'' to build a standard segment tree on range [1,n]:

Node* build(long long l, long long r) {
    Node* x = new(Node);
    if (l == r) return x;
    long long mid = (l + r) / 2;
    x -> lchild = build(l, mid);
    x -> rchild = build(mid + 1, r);
    return x;
}

Chenjb submitted his code, but unfortunately, got MLE (Memory Limit Exceeded). Soon Chenjb realized that his program will new a large quantity of nodes, and he decided to reduce the number of nodes by pruning:

Node* build(long long l, long long r) {
    Node* x = new(Node);
    if (r - l + 1 <= k) return x;
    long long mid = (l + r) / 2;
    x -> lchild = build(l, mid);
    x -> rchild = build(mid + 1, r);
    return x;
}

You know, Chenjb is a freshman, so he will try different values of k to find the optimal one. You will be given the values of n and k, please tell him the number of nodes that will be generated by his new program.

Input

The first line contains a single integer T (1≤T≤10000), the number of test cases. For each test case:

The only line contains two integers n and k (1≤k≤n≤1018), denoting a query.

Output

For each query, print a single line containing an integer, denoting the number of segment tree nodes.

Sample Input

3
100000 1
100000 50
1000000000000000000 1

Sample Output

199999
4095
1999999999999999999

sb了，一开始想能不能打表推式子，后来想到了记忆化又不知道咋写...

可以发现，给定一段区间[l, r]，这个区间中的能生成的节点数和左右端点大小无关，只和r - l + 1有关，且不同的区间长度并不是很多，因此考虑记录长度对应的生成节点数。

#include <bits/stdc++.h>
using namespace std;
#define int long long
long long n, k;
long long cnt  = 0;
map<long long, long long> mp;
long long ans = 0;
long long dfs(long long l, long long r) {
	cnt++;
	if(r - l + 1 <= k) {
		return 1;
	}
	if(mp.find(r - l + 1) != mp.end()) {
		long long tmp = mp[r - l + 1];
		return tmp;
	}
	long long tmp = 1;
	long long mid = (l + r) >> 1;
	tmp += dfs(l, mid);
	tmp += dfs(mid + 1, r);
	mp[r - l + 1] = tmp;
	return tmp;
}
signed main() {
	int t;
	cin >> t;
	while(t--) {
		cnt = 0;
		ans = 0;
		mp.clear();
		cin >> n >> k;
		cout << dfs(1, n) << endl;
	}
	return 0;
}