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

#include<stdio.h>
#include<vector>
#include<math.h>
#include<cmath>
using namespace std;

int getGCD(long long a,long long b)
{
while(b!=0)
{
int t = a % b;
a = b ;
b = t ;
}

return a;
}
int main()
{
int n,i;
long long tem1,tem2;
scanf("%d",&n);
vector<long long> child,mother;
for(i = 0; i < n ;i ++)
{

scanf("%lld/%lld",&tem1,&tem2);
child.push_back(tem1);
mother.push_back(tem2);
}

long long MonSum = 1;
long long ChildSum = 0;
long long tem ;
for(i = 0; i < mother.size() ;i++)
{
ChildSum = mother[i]*ChildSum + MonSum*child[i];
MonSum *= mother[i];
tem = getGCD(abs(MonSum) , abs(ChildSum));
ChildSum = ChildSum / tem;
MonSum = MonSum / tem;
}


long long pre;

pre = ChildSum/MonSum;
ChildSum = ChildSum % MonSum;
tem = getGCD(abs(MonSum) , abs(ChildSum));
ChildSum = ChildSum / tem;
MonSum = MonSum / tem;
if(ChildSum == 0)
{
printf("%lld
",pre);    
}
else if( ChildSum != 0 && pre == 0)
{
printf("%lld/%lld
",ChildSum,MonSum);
}
else if( ChildSum != 0 && pre != 0)
{
printf("%lld ",pre);    
printf("%lld/%lld
",ChildSum,MonSum);
}


return 0;
}