L1-009 N个数求和
本题的要求很简单，就是求N个数字的和。麻烦的是，这些数字是以有理数分子/分母的形式给出的，你输出的和也必须是有理数的形式。
易错点
浮点错误：除数或者取余为0运行超时：求最小公倍数用递归会超时
#include<iostream>
using namespace std;
long int ele1 = 0, dom1 = 0;
//进行约分
void yuefen(long int& a, long int& b) {
long int max = a > b ? a : b;
long int min = a > b ? b : a;
long int c = 1;
//当分子为0时候
if (min == 0)
a = 0;
else {
c = max % min;
while (c != 0) {
max = min;
min = c;
c = max % min;
}// min
a = a / min;
b = b / min;
}
}
//jieguo
void jieguo(long int a, long int b) {
if (a == 0)
cout << 0;
else if (a / b == 0 )
cout << a << "/" << b << endl;
else if (a % b == 0)
cout << a / b << endl;
else
cout << a / b << " " << a % b << "/" << b << endl;
}
void add(long int ele, long int dom, int i, int num) {
//long int ele1 = 0, dom1 = 0;
if (i == 0)
ele1 = ele, dom1 = dom;
else {
long int mutiple = dom1* dom;//最小公倍数用递归会超时
long int mutiple_ele = (mutiple / dom1) * ele1 + (mutiple / dom) * ele;
yuefen(mutiple_ele, mutiple);//yuefen finished
ele1 = mutiple_ele; dom1 = mutiple;
if (i == num - 1)
jieguo(ele1, dom1);
}
}
int main() {
int num; cin >> num;
char c;
long int ele;
long int dom;
for (int i = 0; i < num; i++) {
cin >> ele >> c >> dom;
add(ele, dom, i, num);
}
return 0;
}
