PAT 1081. Rational Sum (20) GCD

1081. Rational Sum (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -1/6 1/8

Sample Output 3:

7/24

懒得写了，直接贴别人的了。不过这个算法没考虑负号居然也能过zzzzz

求这些分数的和，并且最后的形势是真分数形式；比如  6/3 =2; 4/8=1/2;10/3=3 1/3;
输入的可能是假分数形式，先把假分数的处理成真分数在进行加减，否则会超时；
用到辗转相除法求最大公约数   被除数÷除数=商……余数；
                                      如果余数是0，那么此时的被除数是最大公约数；
                                             否则 被除数=除数；除数=余数，继续
int gcd(int b1, int b2){     return b2==0 ? b1 : gcd(b2, b1%b2);}

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月17日 14:32答案正确201081C++ (g++ 4.7.2)1436datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确14364/41答案正确14364/42答案正确11804/43答案正确13844/44答案正确13084/4

#include<iostream>   using namespace std;   int gcd(int b1, int b2){     return b2==0 ? b1 : gcd(b2, b1%b2);}int main(){  int N;  long int a1, a2, b1, b2,Gmax;  char ctemp;  cin >> N ;  a1 = 0;  b1 = 1;  while (N--)  {    cin >> a2 >> ctemp >> b2;      Gmax =gcd(b2, a2);    b2 /= Gmax;    a2 /= Gmax;    Gmax = gcd(b1, b2);    b2 /= Gmax;    a1 = a1*b2+a2*b1/Gmax;    b1 = b1*b2;   }     if(a1%b1 == 0)cout << a1 / b1 << endl;  else  {      Gmax = gcd(b1, a1);    b1 /= Gmax;    a1 /= Gmax;    if (0 == a1 / b1)    {      cout << a1 << "/" << b1 << endl;    }    else    {      cout << a1 / b1 << " " << a1%b1 << "/" << b1 << endl;    }  }    system("pause");  return 0;}