hdoj2057A + B Again（字符串）

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 12

1A -9

-1A -12

1A -AA

Sample Output

0

2C

11

-2C
-90

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
char a[101],b[101],c[101],d[101],zong[101];
while(~scanf("%s %s",a,b))
{
__int64 sum,sum1=0,sum2=0;
int n1=strlen(a),n2=strlen(b);
for(int i=0;i<n1;i++)
{
if(a[i]=='+'||a[i]=='-')
continue;
if(a[i]=='A')
a[i]=10;
else if(a[i]=='B')
a[i]=11;
else if(a[i]=='C')
a[i]=12;
else if (a[i]=='D')
a[i]=13;
else if(a[i]=='E')
a[i]=14;
else if(a[i]=='F')
a[i]=15;
if(a[0]!='-')
{
if(a[i]>=48)
a[i]-='0';
sum1*=16;
sum1+=a[i];
}
else {
if(a[i]>=48)
a[i]-='0';
sum1*=16;
sum1-=a[i];
}
}
for(int i=0;i<n2;i++)
{
if(b[i]=='+'||b[i]=='-')
continue;
if(b[i]=='A')
b[i]=10;
else if(b[i]=='B')
b[i]=11;
else if(b[i]=='C')
b[i]=12;
else if(b[i]=='D')
b[i]=13;
else if(b[i]=='E')
b[i]=14;
else if(b[i]=='F')
b[i]=15;
if(b[0]!='-')
{
if(b[i]>=48)
b[i]-='0';
sum2*=16;
sum2+=b[i];
}
else {
if(b[i]>=48)
b[i]-='0';
sum2*=16;
sum2-=b[i];
}
}
sum=sum1+sum2;
if(sum<0)
{
printf("-");
sum=-sum;
}
if(sum==0)
printf("0\n");
else
{
int k=0;
while(sum>0)
{
if(sum%16==10)
{
zong[k++]='A';
sum/=16;
}
else if(sum%16==11)
{
zong[k++]='B';
sum/=16;
}
else if(sum%16==12)
{
zong[k++]='C';
sum/=16;
}
else if(sum%16==13)
{
zong[k++]='D';
sum/=16;
}
else if(sum%16==14)
{
zong[k++]='E';
sum/=16;
}
else if(sum%16==15)
{
zong[k++]='F';
sum/=16;
}
else
{
zong[k++]=sum%16+'0';
sum/=16;
}
}
for(int i=k-1;i>=0;i--)
printf("%c",zong[i]);
printf("\n");
}
}
return 0;
}思路：这道字符串wa了好多次，最开始做的时候天真的直接以为是输出a+b，然后wa，后来以为是定义型小了，用__int64 还是wa，最后才发现是用字符组做。这道题是16进制的数，可以先把他转化为十进制加减后在转化回十六进制输出。注意判断正负号，如果是负号可以先保存下然后用正值计算。最后结果如果是负数可以先直接输出-号然后在输出转化后的数。