洛谷P1291 [SHOI2002]百事世界杯之旅

题目链接：

kma

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\n%lld", f1[0], ans1);
	for (register int i = 1; i <= c2; i++) printf("-");
	printf("\n");
	for (register int i = 1; i <= c1; i++) printf(" ");
	printf("%lld", f2[0]);
	return 0;
}