PAT 1081 Rational Sum[分子求和][比较]
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. 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
(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. 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
题目大意:给出几个分数求和。
//又是分数求和的题目,细节会很多啊。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <map> using namespace std; int n; int nums[100]; int getSum(int s,int f){ if(f>=n){//如果已经到了尽头,那么就对s约分并且输出 //还是需要先根据分母找出所有的最小公倍数呢? //一点也不会。写不下去。 } } int main() { cin>>n; for(int i=0;i<n;i++){ cin>>nums[i]; } cout<<getSum(0,1); return 0; }
代码转自:https://www.liuchuo.net/archives/2108
#include <iostream> #include <cstdlib> 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 n, a, b, suma = 0, sumb = 1, gcdvalue;//注意数据类型的定义。 scanf("%lld", &n); for(int i = 0; i < n; i++) { scanf("%lld/%lld", &a, &b); gcdvalue = gcd(a, b);//找到最大公约数进行约分。 a = a / gcdvalue; b = b / gcdvalue; suma = a * sumb + suma * b;//分子 sumb = b * sumb;//计算分母 gcdvalue = gcd(suma, sumb); sumb = sumb / gcdvalue;//再约分。 suma = suma / gcdvalue; } long long integer = suma / sumb; suma = suma - (sumb * integer);//计算余下的分子。 if(integer != 0) { printf("%lld", integer); if(suma != 0) printf(" ");//如果余下的分子不为0. } if(suma != 0) printf("%lld/%lld", suma, sumb); if(integer == 0 && suma == 0) printf("0"); return 0; }
//学习了,需要经常复习吧,要不还是不会。