Simple operation of Oj-244-16 in Nanyang

Simple operation time limit for 16 binary:MS | Memory limit:65535 KB Difficulty:1

Describe

now I'm going to give you a 16-in-a-plus-minus expression that requires the result of a 8-input expression.

Input

The

first line enters a positive integer T (0<t<100000)
Next there is t line, each line input a string s (length less than 15) string has two numbers and a plus or a minus, and the expression is legal and all operations are less than 31 bits

Output

Each expression output takes one row, and the result of the output expression 8 binary.

Sample input

329+482318be+67844ae1-3d6c

Sample output

441141001026565


I want to take this problem to practice the conversion of the system, read the best code to know that C language has a direct hexadecimal, octal placeholder
My Code

#include <iostream>#include<cstdio>using namespacestd;intFdfCharc) {    if(c >='a')        returnC-'a'+Ten; Else        returnC-'0';}intMain () {intN; CharC; scanf ("%d",&N);    GetChar ();  while(n--)    {        intA =0, B =0, d =0, sign =-1;  while(1{//a is the core of the binary-to-10 binary is the d=x*a^n+y*a^n-1+..................+z*a^0 scanf ("%c",&c); if(c = ='+'|| c = ='-')                 Break; A= A * -+HD (c); }        if(c = ='+') Sign=1; Else Sign=0;  while(1) {scanf ("%c",&c); if(c = ='\ n')                 Break; b= b* -+HD (c); }        if(sign) d = A +b; ElseD = A-b; printf ("%o\n", D); }}

Best code:

 on. #include <stdio.h> Geneva.intMain ()Geneva. {Geneva.intT; to. scanf ("%d",&T); .. while(t--) -. { ,.inta,b,d; the.CharC;Ten. scanf ("%x%c%x",&a,&c,&b); One.if(c=='+') d=a+b; A.Elsed=a-b; -.if(d>=0) -. printf ("%o\n", d); the.Elseprintf"-%o\n",-d); -.} -.}

Simple operation of Oj-244-16 in Nanyang