UVA 10288 - Coupons（数学期望+分数计算）

题目链接 https://cn.vjudge.net/problem/UVA-10288

【题意】
有n种不同的彩票，如果搜集齐所有的彩票就可以兑换大奖（n<=33）而每次买到的彩票种类是随机的，问你在平均情况下，需要买多少张彩票才能兑换大奖呢？用分数表示答案

【思路】
已经有 k$k$ 种彩票，设 s=kn$s=\frac{k}{n}$ ，那么假设再买到一种新的彩票需要的次数为 t$t$ 次，对应的概率为 st−1(1−s)$s^{t-1}(1-s)$，所以平均需要的次数为

C(k)=∑t=1+∞st−1(1−s)t=(1−s)∑t=1+∞st−1t$$C(k)=\sum _{t=1}^{+\mathrm{\infty }}{s}^{t-1}(1-s)t=(1-s)\sum _{t=1}^{+\mathrm{\infty }}{s}^{t-1}t$$

然后用错位相减法，设

F(t)=∑t=1+∞st−1t=1+2s+3s2+4s3+......$$F(t)=\sum _{t=1}^{+\mathrm{\infty }}{s}^{t-1}t=1+2s+3s^2+4s^3+......$$

sF(t)=∑t=1+∞stt=s+2s2+3s3+4s4+......$$sF(t)=\sum _{t=1}^{+\mathrm{\infty }}{s}^{t}t=s+2s^2+3s^3+4s^4+......$$

则

F(t)−sF(t)=1+s+s2+s3+...=limn→+∞1−sn+11−s=11−s$$F(t)-sF(t)=1+s+s^2+s^3+...=\underset{n\to +\mathrm{\infty }}{lim}\frac{1-s^{n+1}}{1-s}=\frac{1}{1-s}$$

所以

F(t)=1(1−s)2$$F(t)=\frac{1}{(1-s{)}^2}$$

C(k)=11−s=nn−k$$C(k)=\frac{1}{1-s}=\frac{n}{n-k}$$

最后的结果即为

∑k=0n−1C(k)=n∑k=0n−11n−k$$\sum _{k=0}^{n-1}C(k)=n\sum _{k=0}^{n-1}\frac{1}{n-k}$$

结果需要用分数表示，这里存一个分数计算的板子

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

long long __gcd(long long a,long long b){
    if(b==0) return a;
    return __gcd(b,a%b);
}

struct fraction {
    long long numerator;//分子
    long long denominator;//分母
    fraction() {
        numerator = 0;
        denominator = 1;
    }
    fraction(long long num) {
        numerator = num;
        denominator = 1;
    }
    fraction(long long a, long long b) {
        numerator = a;
        denominator = b;
        this->reduction();
    }

    void operator = (const long long num) {
        numerator = num;
        denominator = 1;
        this->reduction();
    }

    void operator = (const fraction &b) {
        numerator = b.numerator;
        denominator = b.denominator;
        this->reduction();
    }

    fraction operator + (const fraction &b) const {
        long long gcdnum = __gcd(denominator, b.denominator);
        fraction tmp = fraction(numerator*(b.denominator/gcdnum) + b.numerator*(denominator/gcdnum), denominator/gcdnum*b.denominator);
        tmp.reduction();
        return tmp;
    }

    fraction operator + (const int &b) const {
        return ((*this) + fraction(b));
    }

    fraction operator - (const fraction &b) const {
        return ((*this) + fraction(-b.numerator, b.denominator));
    }

    fraction operator - (const int &b) const {
        return ((*this) - fraction(b));
    }

    fraction operator * (const fraction &b) const {
        fraction tmp = fraction(numerator*b.numerator, denominator * b.denominator);
        tmp.reduction();
        return tmp;
    }

    fraction operator * (const int &b) const {
        return ((*this) * fraction(b));
    }

    fraction operator / (const fraction &b) const {
        return ((*this) * fraction(b.denominator, b.numerator));
    }

    void reduction() {
        if (numerator == 0) {
            denominator = 1;
            return;
        }
        long long gcdnum = __gcd(numerator, denominator);
        numerator /= gcdnum;
        denominator /= gcdnum;
    }
};

fraction a[50];

int main(){
    int n;
    while(scanf("%d",&n)==1){
        for(int i=1;i<=n;++i){
            a[i]=fraction(1,i);
        }
        for(int i=2;i<=n;++i){
            a[1]=a[1]+a[i];
        }
        a[1]=a[1]*n;
        long long x=a[1].numerator;
        long long y=a[1].denominator;
        if(y==1) printf("%lld\n",x);
        else{
            long long tmp=x/y,tmpy=y;
            int ctmp=0,cy=0;
            while(tmp){
                ++ctmp;
                tmp/=10;
            }
            while(tmpy){
                ++cy;
                tmpy/=10;
            }
            for(int i=0;i<=ctmp;++i) putchar(' ');
            printf("%lld\n",x%y);
            printf("%lld ",x/y);
            for(int i=0;i<cy;++i) putchar('-');
            puts("");
            for(int i=0;i<=ctmp;++i) putchar(' ');
            printf("%lld\n",y);
        }
    }
    return 0;
}