Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum.
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.
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.
5 2/5 4/15 1/30 -2/60 8/3
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[111]; int main(){ long long a=0,b=1; //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<0)?(-aa):aa,bb); a = aa/g; b = bb/g; } long long x = a/b,y = a%b; if(y == 0) printf("%lld\n",x); else{ if(x) printf("%lld ",x); printf("%lld/%lld\n",y,b); } return 0; }
原文：https://www.cnblogs.com/cunyusup/p/10663460.html