1088. Rational Arithmetic (20)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
提交代码
分情况讨论。其实代码可以写的更加简洁的，很多过程是重复的。
1 #include<cstdio>
2 #include<cstring>
3 #include<stack>
4 #include<algorithm>
5 #include<iostream>
6 #include<stack>
7 #include<set>
8 #include<map>
9 #include<vector>
10 using namespace std;
11 long long gcd(long long a,long long b){//不能保证返回的符号，但能保证返回的绝对值大小
12 if(b==0){
13 if(a<0){
14 a=-a;
15 }
16 return a;
17 }
18 return gcd(b,a%b);
19 }
20 long long com;
21 void output(long long fz1,long long fm1){
22 if(fz1==0){
23 printf("0");
24 return;
25 }
26 com=gcd(fz1,fm1);
27 fz1/=com;
28 fm1/=com;
29
30 //cout<<fz1<<" "<<fm1<<" "<<com<<endl;
31
32 if(fz1%fm1==0){
33 printf("%lld",fz1/fm1);
34 }
35 else{
36 if(fz1/fm1){
37 printf("%lld ",fz1/fm1);
38 if(fz1<0){
39 fz1=-fz1;
40 }
41 }
42 printf("%lld/%lld",fz1-fz1/fm1*fm1,fm1);
43 }
44 }
45 void add(long long fz1,long long fm1,long long fz2,long long fm2){
46 if(fz1<0){
47 printf("(");
48 output(fz1,fm1);
49 printf(")");
50 }
51 else{
52 output(fz1,fm1);
53 }
54 printf(" + ");
55 if(fz2<0){
56 printf("(");
57 output(fz2,fm2);
58 printf(")");
59 }
60 else{
61 output(fz2,fm2);
62 }
63 com=gcd(fm1,fm2);
64 fz2*=fm1/com;
65 fz1*=fm2/com;
66 fm1*=fm2/com;
67
68 //cout<<fz1<<" "<<fm1<<" "<<fz2<<" "<<fm2<<endl;
69
70
71 printf(" = ");
72 fz1+=fz2;
73 if(fz1<0){
74 printf("(");
75 output(fz1,fm1);
76 printf(")");
77 }
78 else{
79 output(fz1,fm1);
80 }
81 }
82 void sub(long long fz1,long long fm1,long long fz2,long long fm2){
83 if(fz1<0){
84 printf("(");
85 output(fz1,fm1);
86 printf(")");
87 }
88 else{
89 output(fz1,fm1);
90 }
91 printf(" - ");
92 if(fz2<0){
93 printf("(");
94 output(fz2,fm2);
95 printf(")");
96 }
97 else{
98 output(fz2,fm2);
99 }
100 com=gcd(fm1,fm2);
101 fz2*=fm1/com;
102 fz1*=fm2/com;
103 fm1*=fm2/com;
104 printf(" = ");
105 fz1-=fz2;
106 if(fz1<0){
107 printf("(");
108 output(fz1,fm1);
109 printf(")");
110 }
111 else{
112 output(fz1,fm1);
113 }
114 }
115 void mul(long long fz1,long long fm1,long long fz2,long long fm2){
116 if(fz1<0){
117 printf("(");
118 output(fz1,fm1);
119 printf(")");
120 }
121 else{
122 output(fz1,fm1);
123 }
124 printf(" * ");
125 if(fz2<0){
126 printf("(");
127 output(fz2,fm2);
128 printf(")");
129 }
130 else{
131 output(fz2,fm2);
132 }
133 printf(" = ");
134 fz1*=fz2;
135 fm1*=fm2;
136 if(fz1<0){
137 printf("(");
138 output(fz1,fm1);
139 printf(")");
140 }
141 else{
142 output(fz1,fm1);
143 }
144 }
145 void quo(long long fz1,long long fm1,long long fz2,long long fm2){
146 if(fz1<0){
147 printf("(");
148 output(fz1,fm1);
149 printf(")");
150 }
151 else{
152 output(fz1,fm1);
153 }
154 printf(" / ");
155 if(fz2<0){
156 printf("(");
157 output(fz2,fm2);
158 printf(")");
159 }
160 else{
161 output(fz2,fm2);
162 }
163 printf(" = ");
164 fz1*=fm2;
165 fm1*=fz2;
166 if(fm1==0){
167 printf("Inf");
168 return;
169 }
170 if(fm1<0){
171 fm1=-fm1;
172 fz1=-fz1;
173 }
174 if(fz1<0){
175 printf("(");
176 output(fz1,fm1);
177 printf(")");
178 }
179 else{
180 output(fz1,fm1);
181 }
182 }
183 int main()
184 {
185 //freopen("D:\\INPUT.txt","r",stdin);
186 long long fz1,fm1,inter1,fz2,fm2,inter2;
187 scanf("%lld/%lld %lld/%lld",&fz1,&fm1,&fz2,&fm2);
188
189 //cout<<fm2<<endl;
190
191 com=gcd(fz1,fm1);
192 fz1/=com;
193 fm1/=com;
194 com=gcd(fz2,fm2);
195
196 //cout<<com<<endl;
197 //cout<<fm2<<endl;
198 fz2/=com;
199 fm2/=com;
200 //cout<<fm2<<endl;
201 //计算
202 //cout<<fz1<<" "<<fm1<<" "<<fz2<<" "<<fm2<<endl;
203 add(fz1,fm1,fz2,fm2);
204 printf("\n");
205 sub(fz1,fm1,fz2,fm2);
206 printf("\n");
207 mul(fz1,fm1,fz2,fm2);
208 printf("\n");
209 quo(fz1,fm1,fz2,fm2);
210 printf("\n");
211 return 0;
212 }