題目描述

本題要求編寫程序，計算2個有理數的和、差、積、商。

輸入描述:
輸入在一行中按照“a1/b1 a2/b2”的格式給出兩個分數形式的有理數，其中分子和分母全是整型范圍內的整數，負號只可能出現在分子前，分母不為0。

輸出描述:
分別在4行中按照“有理數1 運算符 有理數2 = 結果”的格式順序輸出2個有理數的和、差、積、商。注意輸出的每個有理數必須是該有理數的最簡形式“k a/b”，其中k是整數部分，a/b是最簡分數部分；若為負數，則須加括號；若除法分母為0，則輸出“Inf”。題目保證正確的輸出中沒有超過整型范圍的整數。

輸入例子:
5/3 0/6

輸出例子:
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

我編寫的本地測試用例

2/3 -4/2
答案：
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

5/3 0/6
答案：
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

0/3 0/6
我的結果：
0 + 0 = 0
0 - 0 = 0
0 * 0 = 0
0 / 0 = Inf

-0/3 -0/6
我的結果：
0 + 0 = 0
0 - 0 = 0
0 * 0 = 0
0 / 0 = Inf

5/3 -4/2
我的結果：
1 2/3 + (-2) = (-1/3)
1 2/3 - (-2) = 3 2/3
1 2/3 * (-2) = (-3 1/3)
1 2/3 / (-2) = (-5/6)

-65535/65534 -1/2
我的結果：
(-1 1/65534) + (-1/2) = (-1 16384/32767)
(-1 1/65534) - (-1/2) = (-16384/32767)
(-1 1/65534) * (-1/2) = 65535/131068
(-1 1/65534) / (-1/2) = 2 1/32767

代碼

#define _CRT_SECURE_NO_WARNINGS #include<iostream> using namespace std; class Rational { public:int up;int down;void input(int up, int down); }; //構造器 void Rational::input(int up, int down) {this->up = up;this->down = down; }//通分 void reduction(Rational* r1, Rational* r2) {int fenMu = r1->down*r2->down;r1->up *= r2->down;r2->up *= r1->down;r1->down = fenMu;r2->down = fenMu;} //格式化輸出 void SimplifyOutput(Rational num) {int up = num.up;//分母int down = num.down;//分子if (up == 0)//如果分母為0則直接輸出{cout << '0';return;}//找最大公因數、約分int factor;//臨時公因數int maxFactor = 1;//最大公因數int min = up > down ? up : down;for (factor = 1; factor <= min; factor++){if (up%factor == 0 && down%factor == 0)//找到公因數{maxFactor = factor;}}up /= maxFactor;down /= maxFactor;//打印if (num.up < 0){cout << "(";//先打印符號if (num.up < 0){up *= (-1);cout << "-";}int integer = up / down;if (integer != 0 && up%down == 0){cout << integer;}else if (integer != 0 && up%down != 0){cout << integer << " ";}if (up%down == 0 && integer != 0)//如果分母為0則不輸出{cout << ")";return;}cout << up % down << "/" << down;cout << ")";}else{int integer = up / down;if (integer != 0){cout << integer << " ";}if (up%down == 0 && integer != 0)//如果分母為0則不輸出{return;}cout << up % down << "/" << down;} } //加法 void add(Rational left, Rational right) {reduction(&left, &right);//通分//計算Rational result;result.up = left.up + right.up;result.down = left.down;//打印SimplifyOutput(left);cout << " + ";SimplifyOutput(right);cout << " = ";SimplifyOutput(result); } //減法 void sub(Rational left, Rational right) {reduction(&left, &right);//通分//計算Rational result;result.up = left.up - right.up;result.down = left.down;//打印SimplifyOutput(left);cout << " - ";SimplifyOutput(right);cout << " = ";SimplifyOutput(result); } //乘法 void multiply(Rational left, Rational right) {//計算Rational result;result.up = left.up*right.up;result.down = left.down*right.down;//打印SimplifyOutput(left);cout << " * ";SimplifyOutput(right);cout << " = ";SimplifyOutput(result); } //除法 void division(Rational left, Rational right) {//計算Rational result;result.up = left.up*right.down;result.down = left.down*right.up;//讓負號始終在分母上if (result.down < 0){result.down *= (-1);result.up *= (-1);}//打印SimplifyOutput(left);cout << " / ";SimplifyOutput(right);cout << " = ";if (right.up == 0)//如果除數是0{cout << "Inf";return;}SimplifyOutput(result); }int main() {Rational left, right;//輸入int a1, a2, b1, b2;scanf("%d/%d %d/%d", &a1, &a2, &b1, &b2);left.up = a1;left.down = a2;right.up = b1;right.down = b2;//加法add(left, right);cout << endl;//減法sub(left, right);cout << endl;//乘法multiply(left, right);cout << endl;//除法division(left, right); }

總結

以上是生活随笔為你收集整理的牛客网_PAT乙级_10234有理数四则运算(20)【通过5/7:格式错误】的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。