UVa10288 Coupons 分数模版

UVa10288

题目非常简单， 答案就是 n/n+n/(n-1)+...+n/1;

要求分数输出。套用分数模板。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 500;   

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;
    }
    void print() {
        if (denominator == 1) printf("%lld\n", numerator);
        else {
            long long num = numerator/denominator;
            long long 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",numerator%denominator);  

            if (num != 0) printf("%lld ", num);
            tmp = denominator;
            while (tmp) {
                printf("-");
                tmp/=10;
            }
            puts("");  

            for (int i = 0; i < len; i++) printf(" ");
            if (len != 0) printf(" ");
            printf("%lld\n",denominator);
        }
    }
} f[maxn];  

int main() {
    int n;
    while (scanf("%d", &n) != EOF) {  

        f[0] = 0;
        for (int i = 1; i <= n; i++)
            f[i] = f[i-1] + fraction(1, i);
        f[n] = f[n] * n;  

        f[n].print();
    }
    return 0;
}