1034 有理数四则运算 (20 分)

1034 有理数四则运算 (20 分)

重点在对分数的处理

1#include <iostream>
2#include <cmath>
3
4using namespace std;
5
6//辗转相除法 求最大公约数 
7int gcd(long long a, long long b){
8    return b == 0 ? a : gcd(b, a % b);
9}
10
11void print(long long a, long long b){
12    long long c = 0; //带分数前面的整数部分，默认是0
13    if(a > 0){ //正数
14        if(b == 1){ //形如3/1
15            printf("%lld", a);
16        }
17        else if(a > b){ //形如5/3
18            c = a / b;
19            a -= b * c;
20            printf("%lld %lld/%lld", c, a, b);
21        }
22        else{ //真分数 形如3/5
23            printf("%lld/%lld", a, b);
24        }
25    }
26    else if(a == 0){ //形如0/3
27        printf("%c", '0');
28    }
29    else{ //负数
30        if(b == 1){ //形如-3/1
31            printf("(%lld)", a);
32        }
33        else if(-1 * a > b){ //形如-5/3
34            c = a / b;
35            a = (-1 * a) % b;
36            printf("(%lld %lld/%lld)", c, a, b);
37        }
38        else{ //真分数
39            printf("(%lld/%lld)", a, b);
40        }
41    }
42}
43
44void add(long long a1, long long b1, long long a2, long long b2){
45    print(a1, b1);
46    printf(" + ");
47    print(a2, b2);
48    printf(" = ");
49    long long a3 = a1 * b2 + a2 * b1;
50    long long b3 = b1 * b2;
51    //化简到最简形式，非带分数形式
52    long long gcd3 = abs(gcd(a3, b3));
53    a3 /= gcd3;
54    b3 /= gcd3;
55    print(a3, b3);
56    printf("\n");
57}
58
59void subtract(long long a1, long long b1, long long a2, long long b2){
60    print(a1, b1);
61    printf(" - ");
62    print(a2, b2);
63    printf(" = ");
64    long long a3 = a1 * b2 - a2 * b1;
65    long long b3 = b1 * b2;
66    //化简到最简形式，非带分数形式
67    long long gcd3 = abs(gcd(a3, b3));
68    a3 /= gcd3;
69    b3 /= gcd3;
70    print(a3, b3);
71    printf("\n");
72}
73
74void multiply(long long a1, long long b1, long long a2, long long b2){
75    print(a1, b1);
76    printf(" * ");
77    print(a2, b2);
78    printf(" = ");
79    long long a3 = a1 * a2;
80    long long b3 = b1 * b2;
81    //化简到最简形式，非带分数形式
82    long long gcd3 = abs(gcd(a3, b3));
83    a3 /= gcd3;
84    b3 /= gcd3;
85    print(a3, b3);
86    printf("\n");
87}
88
89void divide(long long a1, long long b1, long long a2, long long b2){
90    print(a1, b1);
91    printf(" / ");
92    print(a2, b2);
93    printf(" = ");
94    if(a2 == 0){
95        printf("Inf");
96    }
97    else if(a2 < 0){
98        long long a3 = -1 * a1 * b2;
99        long long b3 = -1 * b1 * a2;
100        //化简到最简形式，非带分数形式
101        long long gcd3 = abs(gcd(a3, b3));
102        a3 /= gcd3;
103        b3 /= gcd3;
104        print(a3, b3);
105    }
106    else{
107        long long a3 = a1 * b2;
108        long long b3 = b1 * a2;
109        //化简到最简形式，非带分数形式
110        long long gcd3 = abs(gcd(a3, b3));
111        a3 /= gcd3;
112        b3 /= gcd3;
113        print(a3, b3);
114    }
115    printf("\n");
116}
117
118int main(){
119    long long a1, b1, a2, b2;
120    long long c1 = 0, c2 = 0;
121    scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);
122    //先化简到最简形式，非带分数形式
123    long long gcd1 = abs(gcd(a1, b1));//abs绝对值
124    a1 /= gcd1;
125    b1 /= gcd1;
126    long long gcd2 = abs(gcd(a2, b2));
127    a2 /= gcd2;
128    b2 /= gcd2;
129    //统一用最简形式参与运算
130    add(a1, b1, a2, b2);
131    subtract(a1, b1, a2, b2);
132    multiply(a1, b1, a2, b2);
133    divide(a1, b1, a2, b2);
134    return 0;
135}