This is an A+B Problem

This is an A+B Problem

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

As usual, there will be an A+B problem in warming up, this problem is:
Given two integers A and B, your job is to calculate the sum of A + B.

输入

 There are several test cases, For each test case:
There are two integers A, B for each case (0 ≤ A , B < 10¹⁰⁰⁰).

输出

 For each test case, output one line containing the result of A+B. 

示例输入

1 211111111111 11111111111

示例输出

322222222222

提示

 

来源

 

示例程序

 

#include<stdio.h>    #include<string.h>    char a[2000],b[2000],c[2000];    int main()    {    int i,j,n,m,k,t,l;    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    while(scanf("%s %s",a,b)!=EOF)    {    memset(c,0,sizeof(c));    n=strlen(a);    m=strlen(b);    l=0;    //printf("%s %s\n",a,b);    for(i=n-1,j=m-1;j>=0&&i>=0;i--,j--)    c[l++]=a[i]+b[j]-48;    if(i!=-1)    for(;i>=0;)    c[l++]=a[i--];    if(j!=-1)    for(;j>=0;)    c[l++]=b[j--];        //if(c[l-1]>=58)    //l++;    for(k=0;k<l;k++)    if(c[k]>=58)    {    c[k]=c[k]-10;    c[k+1]+=1;    }    if(c[k]!=0)    {    printf("%c",c[k]+48);    }    for(t=l-1;t>=0;t--)    printf("%c",c[t]);    printf("\n");    }    }