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<stdio.h> 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：
<textarea readonly="readonly" name="code" class="c++"> #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; } </textarea>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！！！
