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, YueGiven 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 fractionalpart 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	20	1081	C++(g++ 4.7.2)	1	436	datrilla

测试点

测试点	结果	用时(ms)	内存(kB)	得d222分/满分
0	答案正确	1	436	4/4
1	答案正确	1	436	4/4
2	答案正确	1	180	4/4
3	答案正确	1	384	4/4
4	答案正确	1	308	4/4

#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;}