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:

 1 /*
 2 Data: 2019-07-05 19:37:00
 3 Problem: PAT_A1081#Rational Sum
 4 AC: 26:24
 5 
 6 题目大意：
 7 给N个分数，求和
 8 */
 9 #include<cstdio>
10 #include<algorithm>
11 using namespace std;
12 const int M = 1e3;
13 struct fr
14 {
15     long long up;
16     long long down;
17 }temp;
18 
19 int gcd(int a, int b)
20 {
21     if(b==0)    return a;
22     else    return gcd(b,a%b);
23 }
24 
25 fr Reduction(fr s)
26 {
27     if(s.up == 0)
28         s.down = 1;
29     else
30     {
31         int d = gcd(abs(s.up), s.down);
32         s.up /= d;
33         s.down /= d;
34     }
35     return s;
36 }
37 
38 fr Add(fr s1, fr s2)
39 {
40     fr s;
41     s.up = s1.up*s2.down+s2.up*s1.down;
42     s.down = s1.down*s2.down;
43     return Reduction(s);
44 }
45 
46 int main()
47 {
48 #ifdef    ONLINE_JUDGE
49 #else
50     freopen("Test.txt", "r", stdin);
51 #endif
52 
53     int n;
54     scanf("%d\n", &n);
55     fr ans = fr{0,1};
56     for(int i=0; i<n; i++)
57     {
58         scanf("%lld/%lld", &temp.up, &temp.down);
59         ans = Add(ans, temp);
60     }
61 
62     if(ans.down==1)
63         printf("%lld\n", ans.up);
64     else if(ans.up >= ans.down)
65         printf("%lld %lld/%lld\n", ans.up/ans.down, abs(ans.up)%ans.down, ans.down);
66     else
67         printf("%lld/%lld\n", ans.up, ans.down);
68 
69     return 0;
70 }