Big vs Big（链表）

【题目描述】

Calculate the addtion of any two positive big integers.

Requirements:

Test data can be more than 64 digits, therefore you MUST use a linked list to store an integer (any big).

【输入】

The first line contains the number of test cases, N.

In the next 2*N lines, each line contains a string of number.

【输出】

Out put N lines. Each line represent the sum of A and B.

【我的程序】

#include <stdlib.h>
#include <iostream>
using namespace std;

typedef struct node{ int num;  node *av; }* wei;

wei newWei(int x)
{
wei y=(wei)malloc(sizeof(node));
y->num=x; y->av=NULL;
return y;
}

int main()
{
int n;
char ch;

cin>> n; cin.get();
for (int i=0;i<n;i++)
{
wei x=newWei(cin.get()-'0');
while ((ch=cin.get())!='\n'){ wei p=newWei(ch-'0'); p->av=x; x=p; }

wei y=newWei(cin.get()-'0');
while ((ch=cin.get())!='\n'){ wei p=newWei(ch-'0'); p->av=y; y=p; }

wei re=newWei(x->num+y->num);
int jw=(re->num)/10; re->num%=10;
x=x->av; y=y->av;

while (x!=NULL && y!=NULL)
{
wei p=newWei(x->num+y->num+jw); jw=(p->num)/10; p->num%=10;
p->av=re; re=p;
x=x->av; y=y->av;
}

while (y!=NULL)
{
wei p=newWei(y->num+jw); jw=p->num/10; p->num%=10;
p->av=re; re=p;
y=y->av;
}

while (x!=NULL)
{
wei p=newWei(x->num+jw); jw=p->num/10; p->num%=10;
p->av=re; re=p;
x=x->av;
}

if (jw>0) cout<< jw;
while (re!=NULL){ cout<< re->num; re=re->av; }
cout<< endl;
}
return 0;
}