1081 Rational Sum (20 分)
阿新 • • 發佈:2018-11-13
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
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
程式碼:
#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; }