(PAT 1088) Rational Arithmetic (分数运算模拟)

Dear 丶 2022-03-18 01:56 180阅读 0赞

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 2/3
    2/3 * (-2) = (-1 1/3)
    2/3 / (-2) = (-1/3)

### Sample Input 2: ###

5/3 0/6

### Sample Output 2: ###

1 2/3 + 0 = 1 2/3
    1 2/3 - 0 = 1 2/3
    1 2/3 * 0 = 0
    1 2/3 / 0 = Inf

解题思路:

模拟分数的运算,比较繁琐的模拟题,代码量较大

我在一个地方卡了比较久,分子分母相同的情况,不能直接输出1,因为可能存在为-1的情况,要输出计算结果

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <string>
    using namespace std;
    struct refs {
    	long long up, down;
    };
    
    long long gcd(long long a, long long b) {
    	if (b == 0) return a;
    	else return gcd(b, a%b);
    }
    
    refs simplify(refs obj) {
    	if (obj.up == 0) {
    		obj.up = 0, obj.down = 1;
    		return obj;
    	}
    	if (obj.down < 0) { //交换负数
    		obj.up = -obj.up;
    		obj.down = -obj.down;
    	}
    	long long gcdNum = gcd(abs(obj.up), abs(obj.down));
    	obj.up /= gcdNum;
    	obj.down /= gcdNum;
    	return obj;
    }
    
    refs sadd(refs obj1,refs obj2) {
    	refs obj;
    	obj.down = obj1.down*obj2.down;
    	obj.up = obj1.up*obj2.down + obj2.up*obj1.down;
    	return simplify(obj);
    }
    
    refs ssub(refs obj1, refs obj2) {
    	refs obj;
    	obj.down = obj1.down*obj2.down;
    	obj.up = obj1.up*obj2.down - obj2.up*obj1.down;
    	return simplify(obj);
    }
    
    refs smulty(refs obj1, refs obj2) {
    	refs obj;
    	obj.down = obj1.down*obj2.down;
    	obj.up = obj1.up * obj2.up;
    	return simplify(obj);
    }
    
    refs sdiv(refs obj1, refs obj2) {
    	refs obj;
    	obj.up = obj1.up * obj2.down;
    	obj.down = obj1.down*obj2.up;
    	return simplify(obj);
    }
    
    void showRes(refs sobj) {
    	if (sobj.up == 0) {
    		printf("0");
    	}
    	else if (sobj.down == 0) {
    		printf("Inf");
    	}
    	else {
    		if (sobj.up < 0) printf("(");
    		if (abs(sobj.up) > abs(sobj.down)) {
    			if (abs(sobj.up) % abs(sobj.down) == 0) {
    				printf("%lld", sobj.up / sobj.down);
    			}
    			else {
    				printf("%lld %lld/%lld", sobj.up / sobj.down, abs(sobj.up % sobj.down), abs(sobj.down));
    			}
    		}
    		else if(abs(sobj.up) == abs(sobj.down)){
    			printf("%lld", sobj.up/sobj.down);
    		}
    		else {
    			printf("%lld/%lld", sobj.up, sobj.down);
    		}
    		if (sobj.up < 0) printf(")");
    	}
    }
    
    int main() {
    	refs input1, input2;
    	scanf("%lld/%lld %lld/%lld", &input1.up, &input1.down, &input2.up, &input2.down);
    	refs obbj[4];
    	char tools[4] = { '+','-','*','/' };
    	int index = 0;
    	obbj[index++]= sadd(input1, input2);
    	obbj[index++] = ssub(input1, input2);
    	obbj[index++] = smulty(input1, input2);
    	obbj[index++] = sdiv(input1, input2);
    	//化简input
    	refs sinput1, sinput2;
    	sinput1 = simplify(input1);
    	sinput2 = simplify(input2);
    	
    	for (int i = 0; i < 4; ++i) {
    		showRes(sinput1);
    		printf(" %c ", tools[i]);
    		showRes(sinput2);
    		printf(" = ");
    		showRes(obbj[i]);
    		printf("\n");
    	}
    	
    	system("PAUSE");
    	return 0;
    }