題解

我們把這個函數的遞歸形式畫成一張圖，會發現答案是到每個出度為0的點的路徑的方案數

這個可以用組合數算

記錄一下P[i]為i減幾次PI減到4以內

如果P[i + 1] > P[i]，那麽轉向的路徑走P[i]次，否則走P[i] - 1次

代碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <bitset>
 

#define enter putchar(‘\n‘)
#define space putchar(‘ ‘)
//#define ivorysi
#define pb push_back
#define mo 974711
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define MAXN 1000005
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void
 
 read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < ‘0‘ || c > ‘9‘) {
    if(c == ‘-‘) f = -1;
    c = getchar();
    }
    while(c >= ‘0‘ && c <= ‘9‘) {
    res = res * 10 - ‘0‘ + c;
    c = getchar();
    }
    res = res * f;
}
template<class T>
void out(T x) {
    if
 
(x < 0) {x = -x;putchar(‘-‘);}
    if(x >= 10) out(x / 10);
    putchar(‘0‘ + x % 10);
}
const int MOD = 1000000007;
const db PI = acos(-1.0);
int N,fac[MAXN],inv[MAXN],invfac[MAXN],P[MAXN];
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(mul(fac[n],invfac[m]),invfac[n - m]);
}
void Init() {
    read(N);
    fac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
    inv[1] = 1;
    for(int i = 2 ; i <= N ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    invfac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) invfac[i] = mul(invfac[i - 1],inv[i]);
}
void Solve() {
    for(int i = 4 ; i <= N ; ++i) {
    P[i] = floor((i - 4) / PI) + 1;
    }
    if(N < 4) {out(1);enter;return;}
    int ans = 0;
    for(int i = N ; i >= 3 ; --i) {
    int s = N - i,t = P[i] - (P[i + 1] <= P[i]);
    ans = inc(ans,C(s + t,s));
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}

【51nod】1149 Pi的遞推式