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:

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\n",pre);    

55 }

56 else if( ChildSum != 0 && pre == 0)

57 {

58 printf("%lld/%lld\n",ChildSum,MonSum);

59 }

60 else if( ChildSum != 0 && pre != 0)

61 {

62 printf("%lld ",pre);    

63 printf("%lld/%lld\n",ChildSum,MonSum);

64 }

65 

66 

67 return 0;

68 }