PAT_A1081#Rational Sum

Source:

PAT A1081 Rational Sum (20 分)

Description:

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 (≤), 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 integeris 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

Keys:

• 分数的四则运算

Code:

 /*
 Data: 2019-07-05 19:37:00
 Problem: PAT_A1081#Rational Sum
 AC: 26:24

 题目大意：
 给N个分数，求和
 */
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int M = 1e3;
 struct fr
 {
     long long up;
     long long down;
 }temp;

 int gcd(int a, int b)
 {
     if(b==)    return a;
     else    return gcd(b,a%b);
 }

 fr Reduction(fr s)
 {
     if(s.up == )
         s.down = ;
     else
     {
         int d = gcd(abs(s.up), s.down);
         s.up /= d;
         s.down /= d;
     }
     return s;
 }

 fr Add(fr s1, fr s2)
 {
     fr s;
     s.up = s1.up*s2.down+s2.up*s1.down;
     s.down = s1.down*s2.down;
     return Reduction(s);
 }

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     int n;
     scanf("%d\n", &n);
     fr ans = fr{,};
     for(int i=; i<n; i++)
     {
         scanf("%lld/%lld", &temp.up, &temp.down);
         ans = Add(ans, temp);
     }

     if(ans.down==)
         printf("%lld\n", ans.up);
     else if(ans.up >= ans.down)
         printf("%lld %lld/%lld\n", ans.up/ans.down, abs(ans.up)%ans.down, ans.down);
     else
         printf("%lld/%lld\n", ans.up, ans.down);

     return ;
 }