zoukankan      html  css  js  c++  java
  • Uva 10288 Coupons

    Description

    Coupons in cereal boxes are numbered (1) to (n), and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of (n) coupons?

    Input

    Input consists of a sequence of lines each containing a single positive integer (n),(1 le n le 33), giving the size of the set of coupons. Input is terminated by end of file.

    Output

    For each input line, output the average number of boxes required to collect the complete set of (n) coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of output.

    Sample Input

    2
    5
    17

    Sample Output

    3
    5
    11 --
    12
    340463
    58 ------
    720720

    若当前已有(k)种Coupons,那么获得新的Coupons的概率为(p = frac{n-k}{n}),所以获得一种新Coupons的期望步数为$$p+2p(1-p)+3p(1-p)^2+cdots$$
    用错位相消+无穷等比数列求和数列公式,化简得(frac{n}{n-k}),所以$$ans = nsum_{i = 1}^{n}frac{1}{i}$$

    #include<cstdio>
    #include<cstdlib>
    using namespace std;
    
    typedef long long ll;
    int N;
    
    inline ll gcd(ll a,ll b) { if (!b) return a; return gcd(b,a%b); }
    inline int ws(ll a) { int ret = 0; while (a) a /= 10,++ret; return ret; }
    
    struct node
    {
    	ll a,b,c;
    	inline node() { c = 1; }
    
    	inline void update()
    	{
    		ll g = gcd(b,c); b /= g,c /= g;
    		a += b/c; b %= c;
    	}
    	
    	friend inline node operator + (const node &x,node &y)
    	{
    		node ret; y.update();
    		ret.a = x.a+y.a; ret.c = x.c*y.c/gcd(x.c,y.c);
    		ret.b = x.b*(ret.c/x.c)+y.b*(ret.c/y.c);
    		ret.update(); return ret;
    	}
    
    	friend inline node operator *(const node &x,const int &y)
    	{
    		node ret; ret.a = x.a*y;
    		ll g = gcd(x.c,y); ret.b = (y/g)*x.b; ret.c = x.c/g;
    		ret.update(); return ret;
    	}
    
    	inline void print()
    	{
    		if (!b) printf("%lld
    ",a);
    		else
    		{
    			int t = ws(a);
    			for (int i = t+1;i--;) putchar(' ');
    			printf("%lld
    ",b);
    			printf("%lld ",a);
    			for (int i = ws(c);i--;) putchar('-');
    			puts("");
    			for (int i = t+1;i--;) putchar(' ');
    			printf("%lld
    ",c);
    		}
    	}
    }ans;
    
    int main()
    {
    	freopen("10288.in","r",stdin);
    	freopen("10288.out","w",stdout);
    	while (scanf("%d
    ",&N) != EOF)
    	{
    		ans.a = ans.b = 0; ans.c = 1;
    		for (int i = 1;i <= N;++i)
    		{
    			node tmp; tmp.a = 0,tmp.b = 1,tmp.c = i;
    			ans = ans + tmp;
    		}
    		ans = ans * N; ans.print();
    	}
    	fclose(stdin); fclose(stdout);
    	return 0;
    }
    
  • 相关阅读:
    解决加密PDF文档无法复制文字的问题
    Linux创建用户时让每个用户家目录中自带说明文档
    Linux基础命令cp之拷贝隐藏文件
    Linux基础命令之getent
    Linux用户和组管理命令-用户创建useradd
    Linux用户和组管理命令-用户删除userdel
    Linux用户和组管理命令-用户属性修改usermod
    Linux用户和组管理命令-切换用户su
    Linux-京西百花山
    tee命令
  • 原文地址:https://www.cnblogs.com/mmlz/p/6054844.html
Copyright © 2011-2022 走看看