标签：simple form pre spec problem follow == with out
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
#include<stdio.h> #include<math.h> typedef long long ll; struct Fraction { ll up, down; }f[110],result; ll gcd(ll a, ll b) { if (b == 0) return a; else return gcd(b, a % b); } Fraction Reducation(Fraction result) { //是Fraction!!!不是int啦！！！ if (result.down < 0) { result.up = -result.up; result.down = -result.down; } if (result.up == 0) { result.down = 1; } else { ll d = gcd(abs(result.up), abs(result.down)); result.up /= d; result.down /= d; } return result; } void Output(Fraction result) { if (result.up == 0) { printf("0"); } else if (result.down == 1) { printf("%lld\n", result.up); } else if (result.up > result.down) { printf("%lld %lld/%lld\n", result.up / result.down, result.up % result.down, result.down); } else { printf("%lld/%lld\n", result.up, result.down); } } int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%lld/%lld", &f[i].up, &f[i].down); } result.up = f[0].up; result.down = f[0].down; for (int i = 1; i < n; i++) { result.up = result.up * f[i].down + f[i].up * result.down; result.down = result.down * f[i].down; } result = Reducation(result); Output(result); return 0; }
标签：simple form pre spec problem follow == with out
原文地址：https://www.cnblogs.com/Yaxadu/p/9136593.html
