pat-basic-1034-Rational Number arithmetic

The subject asks to write a procedure, calculates 2 rational number's and, difference, product, quotient.

Input format:

The input gives a rational number in the form of "A1/b1 a2/b2" in a row in two fractions, where the numerator and denominator are all integers within the integer range, and the minus sign may appear only in front of the numerator, and the denominator is not 0.

Output format:

In 4 rows, according to the "rational number 1 operator rational number 2 = result" In the format order Output 2 rational number of the and, difference, product, quotient. Note that each rational number of the output must be the simplest form "K A/b" of the rational number, where k is the integer part, A/b is the simplest fraction, and if it is negative, parentheses are required, and if the division denominator is 0, the "Inf" is output. The title guarantees that the correct output does not exceed integers in the integer range.

Input Sample 1:

2/3 -4/2

Output Example 1:

2/3 + (-2) = (-1 1/3) 2/3-(-2) = 2 2/32/3 * (-2) = (-1 1/3) 2/3/(-2) = ( -1/3)

Input Sample 2:

5/3 0/6

Output Example 2:

1 2/3 + 0 = 1 2/31 2/3-0 = 1 2/31 2/3 * 0 = 2/3/0 = INF

This problem WA for a long time, stuck in ll here ...
The general idea is to-pass after the offer and then turn the number after numerator into a true fraction form.

//ll WA#include <bits/stdc++.h>#defineLL Long Longusing namespacestd;voidCONVERT (ll A, ll b);intMain () {LL A1, B1, A2, B2; scanf ("%lld/%lld%lld/%lld", &AMP;A1, &AMP;B1, &AMP;A2, &B2); //+CONVERT (A1, B1); printf (" + ");    Convert (A2, B2); printf (" = "); CONVERT (A1*B2+A2*B1, b1*B2); printf ("\ n"); //-CONVERT (A1, B1); printf (" - ");    Convert (A2, B2); printf (" = "); CONVERT (A1*B2-A2*B1, b1*B2); printf ("\ n"); //*CONVERT (A1, B1); printf (" * ");    Convert (A2, B2); printf (" = "); CONVERT (A1*A2, b1*B2); printf ("\ n"); // /CONVERT (A1, B1); printf (" / ");    Convert (A2, B2); printf (" = "); if(A2 = =0) {printf ("inf\n"); }    Else{CONVERT (A1*B2, a2*B1); printf ("\ n"); }    return 0;}voidCONVERT (ll A, ll b) {if(A >0&& B <0) {a= -A; b= -b; }    Else if(A <0&& B <0) {a= -A; b= -b; }    //A = B*k    if(a% b = =0){        if(A <0) {printf ("(%LLD)", A/b); }        Else{printf ("%lld", A/b); }        return; }    //otherwise//Simplify     for(inti =2; I <= sqrt (max (b)); ++i) {         while(a% i = =0&& b% i = =0) {a/=i; b/=i; }    }    if(A <0){        //-        if(A/b! =0) {printf ("(%lld%lld/%lld)", A/b, (ABS (a))%B, b); }        Else{printf ("(%LLD/%LLD)", A, b); }    }    Else{        //+        if(A/b! =0) {printf ("%lld%lld/%lld", A/b, a%B, b); }        Else{printf ("%lld/%lld", A, b); }    }}

Capouis ' CODE

pat-basic-1034-Rational Number arithmetic