1081 Rational Sum (20 分) 分数计算+最大公约数

2024-05-16 07:27:39

1081 Rational Sum (20 分)

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:

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

注意：求最大公约数的时候是求分子绝对值和分母绝对值的最大公约数

#include <cstdio>
#include <algorithm>
using namespace std;typedef long long ll;
struct Fenshu{ll up;ll down;
};ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);
}int main(){int n;Fenshu temp,sum;sum.up=0;sum.down=1; scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lld/%lld",&temp.up,&temp.down);sum.up=sum.up*temp.down+temp.up*sum.down;sum.down=sum.down*temp.down;ll d=gcd(abs(sum.up),abs(sum.down));sum.up=sum.up/d;sum.down=sum.down/d;}if(sum.down==1){printf("%lld\n",sum.up);}else if(abs(sum.up)>sum.down){printf("%lld %lld/%lld\n",sum.up/sum.down,abs(sum.up)%sum.down,sum.down);}else{printf("%lld/%lld\n",sum.up,sum.down);}return 0;}

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