LOJ2803 CCC2018 平衡树 数论分块、记忆化搜索

传送门

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题意差评，其实就是一个递推式：(f_1 = 1 , f_i = sumlimits_{j=2}^i f_{lfloor frac{i}{j} floor})，然后求(f_N)的值

首先(lfloor frac{i}{j} floor)只有(2sqrt{i})种取值，这意味着每一次计算的关联量并不多

而每一次只需要求一组数据的解，于是大力记忆化搜索一下就可以AC了

PS：在Luogu AC此题有263分大礼包QAQ

#include<bits/stdc++.h>
#include<unordered_map>
#define ll long long
//This code is written by Itst
using namespace std;

unordered_map < int , ll > ans;

ll solve(int x){
	if(x == 1) return 1;
	if(ans.find(x) != ans.end()) return ans[x];
	ll sum = 0;
	for(int i = 2 , pi ; i <= x ; i = pi + 1){
		pi = x / (x / i);
		sum += solve(x / i) * (pi - i + 1);
	}
	return ans[x] = sum;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	//freopen("out","w",stdout);
#endif
	int N;
	cin >> N;
	cout << solve(N);
	return 0;
}