本题要求编写程序,计算2个有理数的和、差、积、商。
输入格式:
输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数,其中分子和分母全是整型范围内的整数,负号只可能出现在分子前,分母不为0。
输出格式:
分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式“k a/b”,其中k是整数部分,a/b是最简分数部分;若为负数,则须加括号;若除法分母为0,则输出“Inf”。题目保证正确的输出中没有超过整型范围的整数。
输入样例1:
2/3 -4/2
输出样例1:
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
输入样例2:
5/3 0/6
输出样例2:
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf
解题:简单的分数运算,可能会出错的地方是如果没有用long long可能在第三四的数据点WA,此外此题要用到辗转相除法,否则会在第四个检测点超时。
代码:
<pre name="code" class="cpp">#include<iostream> #include<string> #include<string.h> using namespace std; long long gcd(long a, long b) { if (b == 0) return a; return gcd(b, a%b); } void deal(long long a, long long b) { if (b == 0) cout << "Inf"; else if (a%b == 0) { if (a / b >= 0) { cout << a / b; } else cout << "(" << a / b << ")"; } else { long long fu = a / b; a -= fu*b; long long m = a, n = b; if (m < 0) m = 0 - m; if (n < 0) n = 0 - n; long long s = gcd(m, n); a /= s; b /= s; if (fu < 0) { if (a < 0) a = 0 - a; cout << "(" << fu << " " << a << "/" << b << ")"; } else if (fu != 0) { if (a < 0) a = 0 - a; cout << fu << " " << a << "/" << b; } else { if (b < 0) { b = 0 - b; a = 0 - a; } if (a < 0) { cout << "(" << a << "/" << b << ")"; } else cout << a << "/" << b; } } } int main() { char q = 0; long long a, b, c, d; cin >> a >> q >> b >> c >> q >> d; long long jin1 = 0, jin2 = 0; char s[4] = { '+','-','*','/' }; for (long long i = 0; i< 4; i++) { deal(a, b); cout << " " << s[i] << " "; deal(c, d); cout << " = "; switch (i) { case 0: deal(a*d + c*b, b*d); cout << endl; break; case 1: deal(a*d - c*b, b*d); cout << endl; break; case 2: deal(a*c, b*d); cout << endl; break; case 3: if (c < 0) { a = 0 - a; c = 0 - c; } deal(a*d, b*c); break; } } }