HDU 2057 A+B Again 手动解决办法

主要是坑在了空间释放上QAQ，每次gets完之后进行处理，处理完之后必须得重新初始化字符串！！！！

真的是大坑，，一道水题做了一晚上……

题目如下：

A+B Again

Problem Description

There must be many A + B problems in our HDOJ , now a new one is coming.
Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too.
Easy ? AC it !

 

Input

The input contains several test cases, please process to the end of the file.
Each case consists of two hexadecimal integers A and B in a line seperated by a blank.
The length of A and B is less than 15.

 

Output

For each test case,print the sum of A and B in hexadecimal in one line.

 

Sample Input

+A -A+1A 121A -9-1A -121A -AA

 

Sample Output

02C11-2C-90

一般方法是这样的：

#include
int main()
{
    long long n,m,v;
    while(scanf("%llx%llx",&n,&m)==2)
    {
        v=n+m;
        if(v<0)
        {
            v=-v;
            printf("-%llX\n",v);
        }
        else
        printf("%llX\n",v);
    }
     return 0;
 }

我跳的坑是这样的：

 
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
    char s[50]={'\0'};
    long long int x=0,y=0,z=0;
    while(gets(s)!=NULL)
    {
       int l = strlen(s);
       int i=l-1;
       long long int k=1;
       //scanf("%llx%llx",&a,&b);
       char c=s[i];
       //printf("length=%d\n",l);
       while(c!=' ')
       {
           //printf("%c ",c);
           if(c=='-')x=-x;
           if(c!='+' && c!='-'){
                if(c<65)x+=(c-48)*k;
                else if(c<97)x+=(c-55)*k;
                     else x+=(c-87)*k;
           }
           k=k*16;
           i--;
           //printf("%c %llX and",c,x);
           c=s[i];
       }
       k=1;
       //printf("i=%d\n",i);
       i--;
       c=s[i];
      // printf("\n");
       while(i>=0)
       {

           if(c=='-')y=-y;
           if(c!='+' && c!='-'){
                if(c<65)y+=(c-48)*k;
                else if(c<97) y+=(c-55)*k;
                     else y+=(c-87)*k;
           }
           k=k*16;
           i--;
           //printf("%c %llX and",c,y);
           c=s[i];
       }
       //printf("\n");
       z=x+y;
       if(z<0)printf("-%llX\n",-z);
       else printf("%llX\n",z);
       x=0;y=0;z=0;
       for(i=0;i<50;i++)s[i]='\0';
    }
    //free(s);
    return 0;
}

emmm最后还是AC了然而那个大坑是真心不容易跳出来

总结一句，初始化hin重要！！！