HDU 2057 A+B Again 手動解決辦法
阿新 • • 發佈：2018-11-17
主要是坑在了空間釋放上QAQ，每次gets完之後進行處理，處理完之後必須得重新初始化字串！！！！
真的是大坑，，一道水題做了一晚上……
題目如下：
A+B Again
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
一般方法是這樣的：
- #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;
- }
我跳的坑是這樣的：
<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最後還是AC瞭然而那個大坑是真心不容易跳出來
總結一句，初始化hin重要！！！