2019HDU多校第九场 Rikka with Quicksort —— 数学推导&&分段打表

题意

设

$$g_m(n)=\begin{cases}
& g_m(i) = 0,     \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \leq i \leq m\\
& g_m(i) = i-1 + \frac{1}{i}\sum _{j=1}^i(g_m(j) + g_m(i-j)), \ \  i > m\\
\end{cases}$$

现给出 $n$ 和 $m$，求 $g_m(n)$ 模 $1000000007$.

分析

当 $n>m$ 时，易知 $a_n = n-1 + \frac{2}{n} S_{n-1}$，

即 $S_n - S_{n-1} = n-1 + \frac{2}{n} S_{n-1}$，

变形得 $\frac{S_n}{(n+1)(n+2)} = \frac{n-1}{(n+1)(n+2)} + \frac{S_{n-1}}{n(n+1)}$.

令 $b_n = \frac{S_n}{(n+1)(n+2)}$，得 $b_n = \frac{n-1}{(n+1)(n+2)} + b_{n-1}$，

变形 $b_n - b_{n-1} = \frac{3}{n+2} - \frac{2}{n+1} = 2(\frac{1}{n+2}-\frac{1}{n+1}) + \frac{1}{n+2}$，

根据裂项相消得 $b_n - b_m = 2(\frac{1}{n+2} - \frac{1}{m+2}) + \frac{1}{m+3} + \frac{1}{m+4}+...+\frac{1}{n+2}$，

即 $b_n = 2(\frac{1}{n+2} - \frac{1}{m+2}) + S(n+2)-S(m+2)$.

其中 $S(n)$ 表示前 $n$ 个倒数和，可以分段打表。

（ $1e4$ 的表虽然不会超内存限制，但是会超代码长度限制www，最后改成2000

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod = 1000000007;
const int maxn = 1e9 + 10;
const int siz = 500000;  //2000
int f[maxn/siz+10] = {{0,389417116,881884276,553560598,224432428,260751735,117569315,...}  //省略了
ll n, m;

ll qpow(ll a, ll b, ll p)
{
    ll ret = 1;
    while(b)
    {
        if(b&1)  ret = ret * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return ret;
}

ll inv(ll n)
{
    return qpow(n, mod-2, mod);
}

ll S(ll n)
{
    int k = n / siz; //printf("%d\n", k);
    ll tmp = f[k];
    for(int i = k*siz+1;i <= n;i++)  tmp = (tmp + inv(i)) % mod;
    return tmp;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld%lld", &n, &m);
        ll bn = (2*(inv(n+2) - inv(m+2)) + S(n+2) - S(m+2)) % mod;
        ll bn_1 = (2*(inv(n+1) - inv(m+2)) + S(n+1) - S(m+2)) % mod;
        ll sn = (n+1) * (n+2) % mod * bn % mod;
        ll sn_1 = n * (n+1) % mod * bn_1 % mod;
        printf("%lld\n", (sn - sn_1 + 2*mod) % mod);
    }
}

 

 

参考链接：https://blog.csdn.net/baiyifeifei/article/details/99892190