1088 Rational Arithmetic（20 分）
For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient.
Input Specification:
Each input file contains one test case, which gives in one line the two rational numbers in the format a1/b1 a2/b2. The numerators and the denominators are all in the range of long int. If there is a negative sign, it must appear only in front of the numerator. The denominators are guaranteed to be non-zero numbers.
Output Specification:
For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers, respectively. The format of each line is number1 operator number2 = result. Notice that all the rational numbers must be in their simplest form k a/b, where k is the integer part, and a/b is the simplest fraction part. If the number is negative, it must be included in a pair of parentheses. If the denominator in the division is zero, output Inf as the result. It is guaranteed that all the output integers are in the range of long int.
Sample Input 1:
2/3 -4/2
Sample Output 1:
2/3 + (-2) = (-1 1/3)2/3 - (-2) = 2 2/32/3 * (-2) = (-1 1/3)2/3 / (-2) = (-1/3)
Sample Input 2:
5/3 0/6
Sample Output 2:
1 2/3 + 0 = 1 2/31 2/3 - 0 = 1 2/31 2/3 * 0 = 01 2/3 / 0 = Inf
题目大意：给出两个分数，并给出其加减乘除的结果表示，在除法时，如果分子为0，那么输出Inf。
//虽然从来没做过这样的题目，但是感觉还是比较简单的。
#include#include #include using namespace std;int main() { long int a[3],b[3]; long int re1[3],re2[3],re3[3],re[4]; scanf("%ld/%ld %ld/%ld",a[1],a[2],b[1],b[2]); //那么这个就是求最小公倍数，最大公因数的问题了。 long int ming=max(a[2],b[2]); long long x=a[2]*b[2]; for(int i=ming;i<=x;i++){ if(i%a[2]==0&&i%b[2]==0){ ming=i;break; } } a[0]=a[1]/a[2]; b[0]=b[1]/b[2]; a[1]=a[1]*ming/a[2]; b[1]=b[1]*ming/b[2]; a[2]=ming; b[2]=ming; re1[] //需要存好多种形式的结果啊！ return 0;}
//自己写着写着就写不下去了。没有底。各种细节，感觉控制不了。
代码来自:https://www.liuchuo.net/archives/1906
#include#include using namespace std;long long int a, b, c, d;//使用这种方法需要数据的范围很大。long long int gcd(long long int t1, long long int t2) { return t2 == 0 ? t1 : gcd(t2, t1 % t2);}void func(long long int m, long long int n) { int flag1 = 0; int flag2 = 0; if (n == 0) { //分母肯定是不能为0得，如果有1/3 0/1这样的输入，在加法中，分母会变成3而不是0. cout << "Inf"; return ; } if (m == 0) { cout << 0; return ; } if (m < 0) { m = 0 - m; flag1 = 1; } if (n < 0) { n = 0 - n; flag2 = 1; } int flag = 0; if (flag1 == 1 && flag2 == 1) { flag = 0; } else if (flag1 == 1 || flag2 == 1) { flag = 1;//这个主要是来确定整个结果包括分子和分母的符号 } if (m == n) { if (flag == 1) cout << "(-1)";//计算结果为负值，需要加括号 else cout << "1";//计算结果为正值，不需要加。 return; } long long int x = m % n;//因为m不可能=0了，之前已经判断过了，所以此处 //如果x为0，那么肯定就是没有余数。 long long int y = m / n; if (x == 0) { if (flag == 0) cout << y; else cout << "(-" << y << ")"; return ; } else { long long int t1 = m - y * n; long long int t2 = n; long long int t = gcd(t1, t2); t1 = t1 / t; t2 = t2 / t; if (flag == 1) { cout << "(-"; if (y != 0)//假分数 cout << y << " " << t1 << "/" << t2; else cout << t1 << "/" << t2; cout << ")"; } else { //真分数 if (y != 0) cout << y << " " << t1 << "/" << t2; else cout << t1 << "/" << t2; } }}void add() { long long int m, n;//还没见过这个数据类型 m = a * d + b * c;//直接这样得出分子，厉害。 n = b * d;//分母。 func(a, b);//处理方式都是一样的。 cout << " + "; func(c, d); cout << " = "; func(m, n); cout << endl;}void min() { long long int m, n; m = a * d - b * c; n = b * d; func(a, b); cout << " - "; func(c, d); cout << " = "; func(m, n); cout << endl;}void multi() { long long int m, n; m = a * c; n = b * d; func(a, b); cout << " * "; func(c, d); cout << " = "; func(m, n); cout << endl;}void div() { long long int m, n; m = a * d; n = b * c; func(a, b); cout << " / "; func(c, d); cout << " = "; func(m, n); cout << endl;}int main() { scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d); add(); min(); multi(); div(); return 0;}
//学习了！各种细节考虑的都很好。
1.对每一个数都有一个函数进行处理，十分简洁。
2.由于数可能会非常大，所以使用了long long int这种数据类型
3.顺便复习了，如何求两个数得最大公因数。！