UVA 10288

UVA 10288 - Coupons

题目链接

题意：n个张票，每张票取到概率等价，问连续取一定次数后，拥有全部的票的期望

思路：递推。f[i]表示还差i张票的时候期望，那么递推式为
f(i)=f(i)∗(n−i)/n+f(i−1)∗i/n+1 化简后递推就可以，输出要输出分数比較麻烦

代码：

#include <cstdio>
#include <cstring>
#include <cmath>

long long gcd(long long a, long long b) {
	if (!b) return a;
	return gcd(b, a % b);
}

long long lcm(long long a, long long b) {
	a = a / gcd(a, b) * b;
	if (a < 0) a = -a;
	return a;
}

struct Fraction {
	long long a, b;
	Fraction() {a = 0; b = 1;}
	
 	Fraction(long long x) {
		a = x; b = 1;
 	}
 	
 	Fraction(long long x, long long y) {
 		a = x; b = y;
  	}
  	
  	void deal() {
  		if (b < 0) {b = -b; a = -a;}
  		long long k = gcd(a, b);
  		if (k < 0) k = -k;
  		a /= k; b /= k;
    }
    
    Fraction operator+(Fraction p) {
    	Fraction ans;
    	ans.b = lcm(b, p.b);
    	ans.a = ans.b / b * a + ans.b / p.b * p.a;
    	ans.deal();
    	return ans;
   	}
   	
   	Fraction operator-(Fraction p) {
    	Fraction ans;
    	ans.b = lcm(b, p.b);
    	ans.a = ans.b / b * a - ans.b / p.b * p.a;
    	ans.deal();
    	return ans;
   	}
   	
   	Fraction operator*(Fraction p) {
   		Fraction ans;
   		ans.a = a * p.a;
   		ans.b = b * p.b;
   		ans.deal();
   		return ans;
   	}
   	
   	Fraction operator/(Fraction p) {
   		Fraction ans;
   		ans.a = a * p.b;
   		ans.b = b * p.a;
   		ans.deal();
   		return ans;
   	}
   	
   	void operator=(int x) {
   		a = x;
   		b = 1;
   	}
   	
   	void print() {
   		if (a == 0) {printf("0
"); return;}
   		if (a % b == 0) {printf("%lld
", a / b); return;}
   		int sn = 0;
   		if (a / b > 0) {
   			long long num = a / b;
      		while (num) {
      			num /= 10;
      			sn++;
      		}
     	}
     	if (sn) for (int i = 0; i <= sn; i++) printf(" ");
     	printf("%lld
", a % b);
     	long long num = b;
     	int cnt = 0;
      	while (num) {
      		num /= 10;
      		cnt++;
      	}
     	printf("%lld ", a / b);
     	for (int i = 0; i < cnt; i++) printf("-");
     	printf("
");
     	for (int i = 0; i <= sn; i++) printf(" ");
     	printf("%lld
", b);
    }
};

Fraction dp[35][35];
int n;

int main() {
	for (long long i = 1; i <= 33; i++) {
		dp[i][0] = 0;
		for (long long j = 1; j <= i; j++)
			dp[i][j] = (dp[i][j - 1] * Fraction(j, i) + Fraction(1)) * Fraction(i, j);
 	}
	while (~scanf("%d", &n)) {
		dp[n][n].print();
 	}
	return 0;
}