PAT 1088.Rational Arithmetic

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

两个分数的加减乘除，输出结果必须化为真分数结果。出了注意计算规则外，没什么特别注意点
  1 #include<stdio.h>
  2 
  3 long long getFactor(long long a, long long b){
  4     if (a<0){
  5         a = -a;
  6     }
  7     if (b<0){
  8         b = -b;
  9     }
 10     if (a < b){
 11         long long temp = b;
 12         b = a;
 13         a = temp;
 14     }
 15     if (b == 0)
 16         return 1;
 17     if (a%b == 0){
 18         return b;
 19     }
 20     long long c = a%b;
 21     return getFactor(b, c);
 22 }
 23 
 24 class Rational{
 25 public:
 26     long long n=0, d;
 27     long long v=0;
 28     void simply(){
 29         long long factor = getFactor(n, d);
 30         n /= factor;
 31         d /= factor;
 32         if (d < 0){
 33             d = -d;
 34             n = -n;
 35         }
 36         n += v*d;
 37         v = 0;
 38         v += n / d;
 39         n -= d * v;
 40     }
 41     void complicate(){
 42         long long factor = getFactor(n, d);
 43         n /= factor;
 44         d /= factor;
 45         n += v*d;
 46         v = 0;
 47     }
 48     void print(){
 49         if (v == 0){
 50             if (n < 0){
 51                 printf("(%lld/%lld)", n, d);
 52             }
 53             else if (n == 0){
 54                 printf("0");
 55             }
 56             else{
 57                 printf("%lld/%lld", n, d);
 58             }
 59             return;
 60         }
 61         if (n == 0){
 62             if (v < 0){
 63                 printf("(%lld)", v);
 64             }
 65             else{
 66                 printf("%lld", v);
 67             }
 68             return;
 69         }
 70         if (v < 0){
 71             printf("(%lld %lld/%lld)", v, -n, d);
 72         }
 73         else{
 74             printf("%lld %lld/%lld", v, n, d);
 75         }
 76     }
 77 };
 78 
 79 void printRes(Rational rat[2], Rational res, char op){
 80     rat[0].print();
 81     printf(" %c ", op);
 82     rat[1].print();
 83     printf(" = ");
 84     res.print();
 85     printf("\n");
 86 }
 87 
 88 int main(void){
 89     Rational rat[2],res;
 90     int i;
 91     scanf("%lld/%lld", &rat[0].n, &rat[0].d);
 92     scanf("%lld/%lld", &rat[1].n, &rat[1].d);
 93     rat[0].simply();
 94     rat[1].simply();
 95     long long factor = getFactor(rat[0].d, rat[1].d);
 96     res.d = rat[0].d / factor * rat[1].d;
 97     res.n = res.d / rat[0].d * rat[0].n + res.d / rat[1].d * rat[1].n;
 98     res.v = rat[0].v + rat[1].v;
 99     res.simply();
100     printRes(rat, res, '+');
101 
102     res.d = rat[0].d / factor * rat[1].d;
103     res.n = res.d / rat[0].d * rat[0].n - res.d / rat[1].d * rat[1].n;
104     res.v = rat[0].v - rat[1].v;
105     res.simply();
106     printRes(rat, res, '-');
107 
108     rat[0].complicate();
109     rat[1].complicate();
110     res.n = rat[0].n*rat[1].n;
111     res.d = rat[0].d*rat[1].d;
112     res.v = 0;
113     res.simply();
114     rat[0].simply();
115     rat[1].simply();
116     printRes(rat, res, '*');
117 
118     rat[0].complicate();
119     rat[1].complicate();
120     res.n = rat[0].n*rat[1].d;
121     res.d = rat[0].d*rat[1].n;
122     res.v = 0;
123     rat[0].simply();
124     rat[1].simply();
125     if (res.d != 0){
126         res.simply();
127         printRes(rat, res, '/');
128     }
129     else{
130         rat[0].print();
131         printf(" / ");
132         rat[1].print();
133         printf(" = Inf\n");
134     }
135     
136     return 0;
137 }

 

posted @ 2018-01-20 13:45  GrayWind  阅读(71)  评论(0)  编辑  收藏  举报