1081. Rational Sum (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

 1 #include<stdio.h>
 2 #include<vector>
 3 #include<math.h>
 4 #include<cmath>
 5 using namespace std;
 6 
 7 int getGCD(long long a,long long b)
 8 {
 9 while(b!=0)
10 {
11 int t = a % b;
12 a = b ;
13 b = t ;
14 }
15 
16 return a;
17 }
18 int main()
19 {
20 int n,i;
21 long long tem1,tem2;
22 scanf("%d",&n);
23 vector<long long> child,mother;
24 for(i = 0; i < n ;i ++)
25 {
26 
27 scanf("%lld/%lld",&tem1,&tem2);
28 child.push_back(tem1);
29 mother.push_back(tem2);
30 }
31 
32 long long MonSum = 1;
33 long long ChildSum = 0;
34 long long tem ;
35 for(i = 0; i < mother.size() ;i++)
36 {
37 ChildSum = mother[i]*ChildSum + MonSum*child[i];
38 MonSum *= mother[i];
39 tem = getGCD(abs(MonSum) , abs(ChildSum));
40 ChildSum = ChildSum / tem;
41 MonSum = MonSum / tem;
42 }
43 
44 
45 long long pre;
46 
47 pre = ChildSum/MonSum;
48 ChildSum = ChildSum % MonSum;
49 tem = getGCD(abs(MonSum) , abs(ChildSum));
50 ChildSum = ChildSum / tem;
51 MonSum = MonSum / tem;
52 if(ChildSum == 0)
53 {
54 printf("%lld
",pre);    
55 }
56 else if( ChildSum != 0 && pre == 0)
57 {
58 printf("%lld/%lld
",ChildSum,MonSum);
59 }
60 else if( ChildSum != 0 && pre != 0)
61 {
62 printf("%lld ",pre);    
63 printf("%lld/%lld
",ChildSum,MonSum);
64 }
65 
66 
67 return 0;
68 }