题目链接：
题目分析：
收集邮票的弱弱弱弱化版，因为是期望，考虑倒推
设(f[i])表示现在已经买齐了(i)种，距离买完它的剩余期望次数
那么下一次抽有(frac{i}{n})的概率抽到已经有的，有(frac{n - i}{n})的概率抽到还没有的
那这两种情况的期望分别是(frac{i}{n} * f[i])和(frac{n - i}{n} * f[i + 1])，再加上它自己的期望(1)
有(f[i] = f[i] * frac{i}{n} + frac{n - i}{n} * f[i + 1] + 1)
化简一下得到(f[i] = f[i + 1] + frac{n}{n - i})
倒回来(dp)即可
输出比较恶心，开两个数组分别记录状态的分子和分母，然后手写一下约分之类的函数
代码：
#include <bits/stdc++.h>
#define N (1000 + 10)
#define int long long
using namespace std;
inline int read() {
int cnt = 0, f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
return cnt * f;
}
int n;
int f1[N], f2[N];
int gcd(int a, int b) {return b ? gcd(b, a % b) : a;}
int lcm(int a, int b) {return a * b / gcd(a, b);}
int calc1(int x, int y, int x_, int y_) {
int LCM = lcm(y, y_);
int d1 = LCM / y, d2 = LCM / y_;
int ans = x * d1 + x_ * d2;
return ans;
}
int calc2(int x, int y, int x_, int y_) {
int LCM = lcm(y, y_);
return LCM;
}
void solve(int &x, int &y) {
int GCD = gcd(x, y);
x /= GCD, y /= GCD;
}
int get_digit(int x) {
int cnt = 0;
while(x) {
++cnt;
x /= 10;
}
return cnt;
}
int ans1, ans2, ans3;
signed main(){
n = read();
f1[n] = 0, f2[n] = 0;
f1[n - 1] = n, f2[n - 1] = 1;
for (register int i = n - 2; ~i; --i) {
f1[i] = calc1(f1[i + 1], f2[i + 1], n, n - i);
f2[i] = calc2(f1[i + 1], f2[i + 1], n, n - i);
solve(f1[i], f2[i]);
}
ans1 = f1[0] / f2[0];
if (f1[0] % f2[0] == 0) return printf("%lld", ans1), 0;
f1[0] = f1[0] % f2[0];
int c1 = get_digit(ans1);
int c2 = get_digit(f2[0]);
for (register int i = 1; i <= c1; i++) printf(" ");
printf("%lld
%lld", f1[0], ans1);
for (register int i = 1; i <= c2; i++) printf("-");
printf("
");
for (register int i = 1; i <= c1; i++) printf(" ");
printf("%lld", f2[0]);
return 0;
}