华为oj 超长正整数相加

解决思路是：将每个字符转换成int类型存放在数组中，然后反向从低位开始相加进位，数组需要反向输出，这样可以保证进位位有存储空间

#include<iostream>
#include<string.h>
#include<stdlib.h>
#define N 10000
using namespace std;

void AddLongInteger(char *a,char *b)
{
int n1 = strlen(a);
int n2 = strlen(b);
int a1
;
int b1
;
int sum
;
memset(a1,0,sizeof(a1));
memset(b1,0,sizeof(b1));
memset(sum,0,sizeof(sum));
int i = n1-1;
int j = n2-1;
int n=0,m=0;
for(;i>=0;i--)
{
a1[n++] =  a[i]-'0';
}
for(;j>=0;j--)
{
b1[m++] =  b[j]-'0';
}
if(n>m)
{
for(i=0;i<n;i++)
{
int temp = a1[i]+b1[i]+sum[i];
/* cout<<"a1[i]"<<a1[i]<<endl;
cout<<"b1[i]"<<b1[i]<<endl;
cout<<"sum[i]"<<sum[i]<<endl;
cout<<"n>m--"<<"n:"<<n<<"m:"<<m<<"i:"<<i<<"temp:"<<temp<<endl;*/
if(temp>=10)
{
sum[i] = temp - 10;
sum[i+1] = 1;
}
else
sum[i] = temp;
}
if(sum
)  cout<<sum
;
i--;
for(;i>=0;i--)
cout<<sum[i];
cout<<endl;
}
else
{     for(i=0;i<m;i++)
{
int temp =  a1[i]+b1[i]+sum[i];
/*cout<<"a1[i]"<<a1[i]<<endl;
cout<<"b1[i]"<<b1[i]<<endl;
cout<<"sum[i]"<<sum[i]<<endl;
cout<<"n<m--"<<"n:"<<n<<"m:"<<m<<"i:"<<i<<"temp:"<<temp<<endl;*/
if(temp>=10)
{
sum[i] = temp - 10;
sum[i+1] = 1;
}
else
sum[i] = temp;
}
if(sum[m])  cout<<sum[m];
i--;
for(;i>=0;i--)
cout<<sum[i];
cout<<endl;
}
}
int main()
{
char  a
;
char  b
;
scanf("%s",&a);
scanf("%s",&b);
AddLongInteger(a,b);
return 0;
}