PAT Rational Sum
Rational Sum (20)
题目描述
Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum.
输入描述:
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.
输出描述:
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.
输入例子:
5
2/5 4/15 1/30 -2/60 8/3
输出例子:
3 1/3
#include<iostream>
#include<cstring>
#include<string>
using namespace std; long long gcd(long long x,long long y){
return y?gcd(y,x%y):x;
} long long lcm(long long x,long long y){
return x/gcd(x,y)*y;
} char s[]; int main(){
long long a=,b=; //a / b
int n;
cin >> n;
while(n--){
cin >> s;
char *t = strstr(s,"/");
if(t) *t = ' '; // a/b + c/d
long long c,d;
sscanf(s,"%lld %lld",&c,&d);
long long aa = a*d + b*c;
long long bb = b*d;
long long g = gcd((aa<)?(-aa):aa,bb);
a = aa/g;
b = bb/g;
}
long long x = a/b,y = a%b;
if(y == ) printf("%lld\n",x);
else{
if(x) printf("%lld ",x);
printf("%lld/%lld\n",y,b);
}
return ;
}
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