1081. Rational Sum (20)

题目链接：https://www.patest.cn/contests/pat-a-practise/1081
Given N rational numbers in the form “numerator/denominator”, you are supposed to calculate their sum.

Input Specification:

Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in the next line N rational numbers “a1/b1 a2/b2 …” where all the numerators and denominators are in the range of “long int”. If there is a negative number, then the sign must appear in front of the numerator.

Output Specification:

For each test case, output the sum in the simplest form “integer numerator/denominator” where “integer” is the integer part of the sum, “numerator” < “denominator”, and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.

Sample Input 1:
5
2/5 4/15 1/30 -2/60 8/3
Sample Output 1:
3 1/3
Sample Input 2:
2
4/3 2/3
Sample Output 2:
2
Sample Input 3:
3
1/3 -1/6 1/8
Sample Output 3:
7/24
注意点：数据范围为int,当两个分母相乘时，最大可以达到long long,所以如果使用int就会溢出，有一个测试点错误

#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const int maxn=110;ll a[maxn],b[maxn];//分别存放分子数组，分母数组 ll gcd(ll x,ll y){//求最大公约数     if(y==0) return x;    else return gcd(y,x%y);}int main(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%lld/%lld",&a[i],&b[i]);    }    ll x=b[0];    for(int i=1;i<n;i++){//求分母数组的最小公倍数         x=x*b[i]/gcd(x,b[i]);       }    for(int i=0;i<n;i++){        a[i]=a[i]*(x/b[i]);    }    ll ans1=0,ans2;    for(int i=0;i<n;i++){        ans1+=a[i];    }    ll t=gcd(abs(ans1),abs(x));     ans1/=t,ans2=x/t;//  printf("%d %d\n",ans1,ans2);    if(ans2==1){        printf("%d\n",ans1);    }else if(abs(ans1)>ans2){        printf("%lld %lld/%lld\n",ans1/ans2,abs(ans1%ans2),ans2);    }else{        printf("%lld/%lld\n",ans1,ans2);    }    return 0;   }