[Project Euler 517]A real recursion 题解

传送门：https://projecteuler.net/problem=517

题意还是很简洁的。

刚开始打了一个表，发现没有什么规律，最后仔细盯了一下式子，发现这是将用1和根号n填，直到大于n-根号n的方案数。

显然这个东西是一个组合数，考虑枚举填了多少个根号，算一下方案即可。单词复杂度是根号的，跑了大概0.4s。

//waz
#include <bits/stdc++.h>

using namespace std;

#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))

typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;

#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))

int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}

const int mod = 1e9 + 7;

int inc(int a, int b) { a += b; return a >= mod ? a - mod : a; }

int dec(int a, int b) { a -= b; return a < 0 ? a + mod : a; }

const int N = 2e7;

int fac[N + 10], rfac[N + 10];

int fpow(int a, int x)
{
	int ret = 1;
	for (; x; x >>= 1)
	{
		if (x & 1) ret = 1LL * ret * a % mod;
		a = 1LL * a * a % mod;
	}
	return ret;
}

int C(int n, int m)
{
	if (m < 0 || n < m) return 0;
	return 1LL * fac[n] * rfac[m] % mod * rfac[n - m] % mod;
}

int work(int n)
{
	int ans = 0;
	double t = sqrt(n);
	for (int i = 0; i * t < n - t; ++i)
	{
		int c = ((n - t) - i * t);
		ans = inc(ans, C(c + i, i));
	}
	for (int i = 0; i < n - t; ++i)
	{
		int c = ((n - t) - i) / t;
		int j = ((n - t) - c * t);
		//for (int k = i; k <= j; ++k) ans = inc(ans, C(c + k, c));
		ans = inc(ans, dec(C(c + j + 1, c + 1), C(c + i, c + 1)));
		//cerr << i << ", " << c << ", " << j << endl;
		i = j;
	}
	return ans;
}

unordered_map<double, int> zz;

double tt;

int work_bf(double n)
{
	if (n < tt) return 1;
	if (zz[n]) return zz[n];
	return zz[n] = inc(work_bf(n - 1), work_bf(n - tt));
}

int bf(int n)
{
	tt = sqrt(n);
	zz.clear();
	return work_bf(n);
}

int main()
{
	for (int i = fac[0] = 1; i <= N; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
	rfac[N] = fpow(fac[N], mod - 2); for (int i = N; i; --i) rfac[i - 1] = 1LL * rfac[i] * i % mod;
	int l, r;
	//cin >> l >> r;
	l = 1e7; r = 1e7 + 1e4;
	int ans = 0, gg = 0;
	for (int i = l + 1; i < r; ++i)
	{
		bool fg = 1;
		for (int j = 2; j * j <= i; ++j)
			if (i % j == 0) fg = 0;
		if (fg) ans = inc(ans, work(i));
	}
	printf("%d
", ans);
	return 0;
}