紫书 例题10-18优惠券 UVa 10288

题意：大街上到处在卖彩票，一元钱一张。购买撕开它上面的锡箔，你会看到一个漂亮的图
案。图案有n种，如果你收集到所有n（n≤33）种彩票，就可以得大奖。请问，在平均情况
下，需要买多少张彩票才能得到大奖呢？如n=5时答案为137/12。
分析：设f[i]代表还有i个优惠券没有收集到的期望开箱次数，那么我们只要计算再没有取到的期望+上一个取到的期望+操作次数1，即f[i]=(n-i)/n*f[i]+i/n*f[i+1]+1,移项后是f[i] = f[i+1] + n/i,然后就是一个分数计算的模板了

整理了一个分数类模板，支持常见操作，记录一下：

const int maxn = 510;typedef long long LL;struct Fraction{    LL a, b; //a代表分子，b代表分母    Fraction(){        a = 0; b = 1;    }    void reduction(){        if(a == 0){            b = 1;            return;        }        LL gcdnum = __gcd(a, b);        a /= gcdnum;        b /= gcdnum;    }    Fraction(LL num){ //整数        a = num; b = 1;    }    Fraction(LL x, LL y){        a = x, b = y;        this->reduction();    }    void operator = (const LL &num){        a = num; b = 1;        this -> reduction();    }    void operator = (const Fraction &y){        a = y.a; b = y.b;        this -> reduction();    }    Fraction operator + (const Fraction &y) const{        LL gcdnum = __gcd(b, y.b);        Fraction tmp = Fraction(a * (y.b / gcdnum) + y.a * (b / gcdnum) , b / gcdnum * y.b);        tmp.reduction();        return tmp;    }    Fraction operator + (const LL &y) const{        return ((*this) + Fraction(y));    }    Fraction operator - (const Fraction &y) const{        return ((*this) + Fraction(-y.a, y.b));    }    Fraction operator - (const LL &y) const{        return ((*this) - Fraction(y));    }    Fraction operator * (const Fraction &y) const{        Fraction tmp = Fraction(a * y.a, b * y.b);        tmp.reduction();        return tmp;    }    Fraction operator / (const Fraction &y) const{        return ((*this) * Fraction(y.b, y.a));    }    void print(){        if(b == 1){            printf("%lld\n", a);        }        else{            LL num = a / b;            LL tmp = num;            int len = 0;            while(tmp){                len++; tmp /= 10;            }            for(int i = 0; i < len; i++){                printf(" ");            }            if(len != 0) printf(" ");            printf("%lld\n", a % b);            if(num != 0) printf("%lld ", num);            tmp = b;            while(tmp){                printf("-");                tmp /= 10;            }            printf("\n");            for(int i = 0; i < len; i++) printf(" ");            if(len != 0) printf(" ");            printf("%lld\n", b);        }    }};

本题完整代码：

#include <bits/stdc++.h>using namespace std;const int maxn = 510;typedef long long LL;struct Fraction{    LL a, b; //a代表分子，b代表分母    Fraction(){        a = 0; b = 1;    }    void reduction(){        if(a == 0){            b = 1;            return;        }        LL gcdnum = __gcd(a, b);        a /= gcdnum;        b /= gcdnum;    }    Fraction(LL num){ //整数        a = num; b = 1;    }    Fraction(LL x, LL y){        a = x, b = y;        this->reduction();    }    void operator = (const LL &num){        a = num; b = 1;        this -> reduction();    }    void operator = (const Fraction &y){        a = y.a; b = y.b;        this -> reduction();    }    Fraction operator + (const Fraction &y) const{        LL gcdnum = __gcd(b, y.b);        Fraction tmp = Fraction(a * (y.b / gcdnum) + y.a * (b / gcdnum) , b / gcdnum * y.b);        tmp.reduction();        return tmp;    }    Fraction operator + (const LL &y) const{        return ((*this) + Fraction(y));    }    Fraction operator - (const Fraction &y) const{        return ((*this) + Fraction(-y.a, y.b));    }    Fraction operator - (const LL &y) const{        return ((*this) - Fraction(y));    }    Fraction operator * (const Fraction &y) const{        Fraction tmp = Fraction(a * y.a, b * y.b);        tmp.reduction();        return tmp;    }    Fraction operator / (const Fraction &y) const{        return ((*this) * Fraction(y.b, y.a));    }    void print(){        if(b == 1){            printf("%lld\n", a);        }        else{            LL num = a / b;            LL tmp = num;            int len = 0;            while(tmp){                len++; tmp /= 10;            }            for(int i = 0; i < len; i++){                printf(" ");            }            if(len != 0) printf(" ");            printf("%lld\n", a % b);            if(num != 0) printf("%lld ", num);            tmp = b;            while(tmp){                printf("-");                tmp /= 10;            }            printf("\n");            for(int i = 0; i < len; i++) printf(" ");            if(len != 0) printf(" ");            printf("%lld\n", b);        }    }} dp[maxn];int main(){    int n;    while(scanf("%d", &n) != EOF)    {        dp[0] = 0;        for(int i = 1; i <= n; i++) dp[i] = dp[i - 1] + Fraction(1, i);        dp[n] = dp[n] * n;        dp[n].print();    }    return 0;}