1081 Rational Sum(20 分)(C++)
阿新 • • 發佈:2019-02-18
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
考點:最簡分數化簡
#include<iostream> #include<cstdio> using namespace std; long long gcd(long long a, long long b) { return b == 0 ? abs(a) : gcd(b, a % b); } int main(){ long long a,b,sum=0,suma=0,sumb=1; int n; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%lld/%lld",&a,&b); if(a/b!=0){ long long x=a/b; sum+=x; a=a%b; } long long p=gcd(b,a); a/=p; b/=p; suma=suma*(b)+a*(sumb); sumb=sumb*b; if(suma/sumb!=0){ long long x=suma/sumb; sum+=x; suma=suma%sumb; } p=gcd(sumb,suma); suma/=p; sumb/=p; } if(sum!=0||suma==0) printf("%lld",sum); if(sum!=0&&suma!=0) printf(" "); if(suma!=0) printf("%lld/%lld",suma,sumb); system("pause"); }