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93 lines
2.6 KiB
C
93 lines
2.6 KiB
C
/**
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* 1034. 有理数四则运算(20)
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*
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* 本题要求编写程序，计算2个有理数的和、差、积、商。
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*
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* 输入格式：
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*
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* 输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数，其中分子和分母全是
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* 整型范围内的整数，负号只可能出现在分子前，分母不为0。
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*
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* 输出格式：
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*
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* 分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、
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* 积、商。注意输出的每个有理数必须是该有理数的最简形式“k a/b”，其中k是整数部分，
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* a/b是最简分数部分；若为负数，则须加括号；若除法分母为0，则输出“Inf”。题目保证正
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* 确的输出中没有超过整型范围的整数。
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*
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* 输入样例1：
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* 2/3 -4/2
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* 输出样例1：
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* 2/3 + (-2) = (-1 1/3)
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* 2/3 - (-2) = 2 2/3
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* 2/3 * (-2) = (-1 1/3)
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* 2/3 / (-2) = (-1/3)
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* 输入样例2：
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* 5/3 0/6
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* 输出样例2：
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* 1 2/3 + 0 = 1 2/3
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* 1 2/3 - 0 = 1 2/3
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* 1 2/3 * 0 = 0
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* 1 2/3 / 0 = Inf
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*/
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#include <stdio.h>
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/* Both parameters take positive value */
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long calcgcd(long a, long b)
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{
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long r;
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while((r = a % b))
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{
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a = b;
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b = r;
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}
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return b;
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}
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/* print a fraction number, giving the numerator and dominator */
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void printfrac(long n, long d)
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{
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if(d == 0) { printf("Inf"); return; }
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/* record the sign and make them positive */
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int inegative = 1;
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if(n < 0) { n = -n; inegative *= -1; }
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if(d < 0) { d = -d; inegative *= -1; }
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/* reduce the fraction */
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long gcd = calcgcd(n, d);
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n /= gcd;
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d /= gcd;
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/* print */
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if(inegative == -1) printf("(-");
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if(n / d && n % d) printf("%ld %ld/%ld", n / d, n % d, d); /* mixed fractions */
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else if(n % d) printf("%ld/%ld", n % d, d); /* proper fractions */
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else printf("%ld", n / d); /* integers */
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if(inegative == -1) printf(")");
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}
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int main()
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{
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long a1, b1, a2, b2;
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scanf("%ld/%ld %ld/%ld", &a1, &b1, &a2, &b2);
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char op[4] = {'+', '-', '*', '/'};
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for(int i = 0; i < 4; i++)
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{
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printfrac(a1, b1); printf(" %c ", op[i]);
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printfrac(a2, b2); printf(" = ");
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switch(op[i])
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{
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case '+': printfrac(a1 * b2 + a2 * b1, b1 * b2); break;
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case '-': printfrac(a1 * b2 - a2 * b1, b1 * b2); break;
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case '*': printfrac(a1 * a2, b1 * b2); break;
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case '/': printfrac(a1 * b2, b1 * a2); break;
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}
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printf("\n");
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}
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return 0;
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}
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