LOJ#6342. 跳一跳(期望)

题意

$n \leqslant 10^5$

sol

随便推一推就好了吧。。

$f[i] = \frac{f[i] + f[i +1] + \dots f[n]}{n - i + 1} + 1$

移一下项，然后化一化，就做完了。。

然而这题卡空间mmp

#include<cstdio>
#include<algorithm>
#include<iostream>
//#define int long long 
#define pair pair<int, int> 
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
using namespace std;
const int maxn = 1e7 + 10, inf = 1e9 + 10, mod = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); 
    return x * f;
}
int n;
int inv[maxn];
main() {
    n = read();
    int s = 0, now = 0;
    inv[1] = 1;
    for(int i = 2; i <= n; i++) inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    for(int i = n - 1; i >= 1; i--) {
        now = 1ll * (s + n - i + 1) * inv[n - i] % mod;
        s = (s + now) % mod; 
    }
    cout << (now + mod) % mod;
    return 0;
}
/*
10000000
*/