Big vs Big（链表）

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【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;}

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