1081 Rational Sum (20 分)
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
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node
{
ll num,den;
}s[105];
int main()
{
ll n, a, b;
scanf("%lld",&n);
scanf("%lld/%lld",&a, &b);
for(int i = 2 ; i <= n; i++)
{
scanf("%lld/%lld",&s[i].num, &s[i].den);
if(b % s[i].den == 0 || s[i].den % b == 0)
{
ll gg = max(b,s[i].den);
ll den = gg;
ll num = gg / b * a + (gg / s[i].den * s[i].num);
a = num;
b = den;
}else
{
ll gg = __gcd(b,s[i].den);
ll den = b * s[i].den / gg;
ll num = den / b * a + den / s[i].den * s[i].num;
b = den;
a = num;
}
ll gg = __gcd(a,b);
a = a / gg;
b = b / gg;
}
ll gg = __gcd(a,b);
a = a / gg;
b = b / gg;
if(a % b == 0)
printf("%lld\n",a / b);
else if(a >= b)
printf("%lld %lld/%lld\n",a / b, a - (a / b * b),b);
else
printf("%lld/%lld\n",a,b);
return 0;
}