题目大意：有$n$个球，每一次取一个球然后放回，问期望多少次取遍所有球

题解：令$f_i$表示已经取了$i$种球，还要取的次数的期望。$f_i=\dfrac in(f_i+1)+\dfrac{n-i}n(f_{i+1}+1),f_n=0$

解个方程可得$f_i=\dfrac n{n-i}+f_{i+1}$，所有答案为$n\sum\limits_{i=1}^n\dfrac 1i$

卡点：无

 

C++ Code：

#include 
     
      #define maxn 10000010const int mod = 20040313;#define mul(x, y) static_cast
      
        (x) * (y) % modinline void reduce(int &x) { x += x >> 31 & mod; }int n, ans;int inv[maxn];int main() {	scanf("%d", &n);	inv[0] = inv[1] = ans = 1;	for (int i = 2; i <= n; ++i) {		inv[i] = mul(mod - mod / i, inv[mod % i]);		reduce(ans += inv[i] - mod);	}	printf("%lld\n", mul(ans, n));	return 0;}