PAT1081Rational Sum

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这道题只要知道最大公约数和最小公倍数的求法就可以做得出来，但是要考虑的输出细节比较繁琐，代码写的很丑啦，不过总算是全部case通过了。

#include <iostream>#include <math.h>#include <stdio.h>using namespace std;struct rational{    int numerator;    int denominator;};int gcd(int a,int b){    if (b==0) {        return a;    }    else        return gcd(b,a%b);}int lcm(int a,int b){    return a*b/gcd(a, b);}rational add(rational sum,rational next){    int l=lcm(sum.denominator, next.denominator);    //开始通分并相加    sum.numerator=(l/sum.denominator)*sum.numerator;    next.numerator=(l/next.denominator)*next.numerator;    sum.numerator+=next.numerator;    sum.denominator=l;    //对结果进行约分    int g=gcd(sum.denominator, sum.numerator);    sum.numerator/=g;    sum.denominator/=g;    return sum;}void output(rational sum){    int integ;    if (sum.numerator==0) {        printf("0\n");        return;    }    if (sum.numerator<0) {        printf("-");        sum.numerator*=-1;    }    integ=sum.numerator/sum.denominator;    if (integ!=0) {        printf("%d",integ);        sum.numerator=sum.numerator-integ*sum.denominator;    }    if (integ!=0&&sum.numerator!=0) {        printf(" ");    }    if (sum.numerator!=0) {        printf("%d/%d\n",sum.numerator,sum.denominator);    }    else        printf("\n");}int main(){    int n;    rational sum,next;   // freopen("/Users/pantingting/Documents/code/data/input", "r", stdin);    scanf("%d",&n);    sum.denominator=1;    sum.numerator=0;    for (int i=0; i<n; i++) {        scanf("%d/%d",&next.numerator,&next.denominator);        sum=add(sum, next);    }    output(sum);    return 0;}

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