A + B Problem II（大数基础加法）

Description
I have a VERY SIMPLE problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
正整数高精度1000位加法
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input

2
1 2
112233445566778899 998877665544332211
Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

题意：基础的大数的加法，将两个数都以字符数组的形式逐一相加，大于10就对10取余然后前一位加1

代码（C）

#include<stdio.h>#include<string.h>int main(){    int t,m,n,a[1010],b[1010],len1,len2;    char str1[1010],str2[1010];    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        memset(a,0,sizeof(a));          memset(b,0,sizeof(b));        scanf("%s%s",str1,str2);        len1=strlen(str1);        len2=strlen(str2);        for(int k=0;k<len1;k++)  //将字符串存入数组里计算           a[len1-1-k]=str1[k]-'0';          for(int k=0;k<len2;k++)             b[len2-1-k]=str2[k]-'0';         if(len1<=len2)  //比较大小后方便得出和的位数        {            t=len1;            len1=len2;            len2=t;        }        for(int j=0;j<len1;j++)        {            a[j]=a[j]+b[j];            if(a[j]>9)            {                a[j]=a[j]%10;                a[j+1]++;            }        }        printf("Case %d:\n",i);        printf("%s + %s = ",str1,str2);        if(a[len1]==0)  //如果两个数相加后并没有进位        {            for(int j=len1-1;j>=0;j--)  printf("%d",a[j]);            printf("\n");        }        else        {            for(int j=len1;j>=0;j--)  printf("%d",a[j]);            printf("\n");        }        if(i!=n)  printf("\n");    }    return 0;}