A + B Problem II

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本文介绍了一种处理大数相加的算法实现方法，通过字符数组存储大整数，并逐位进行加法运算，解决了传统32位整数无法处理的问题。该算法适用于输入的大整数长度不超过1000的情况。

Problem 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.

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

#include<stdio.h>
#include<string.h>
int main()
{
    int n,m;
    scanf("%d",&n);
    m=1;
    while(m<=n){
        char a[1000],b[1000];
        int i,j,c[1000]={0},la,lb,q;
        scanf("%s %s",a,b);
        la=strlen(a)-1;
        lb=strlen(b)-1;
        q=la>lb?la:lb;//比较2数组的长度
        for(i=0;la>=0;i++,la--)
            c[i]=a[la]-'0';
        for(i=0;lb>=0;lb--,i++){
            c[i]+=b[lb]-'0';
            if(c[i]>=10){
                c[i+1]++;
                c[i]-=10;
            }
        }
        while(c[i]!=0){
            if(c[i]>=10){
                c[i+1]++;
                c[i]-=10;
            }
            i++;
        }
        printf("Case %d: ",m);
        printf("%s + %s = ",a,b);
        if(c[q+1]!=0) 
            printf("%d",c[q+1]);//加法最多进1位
        for(q;q>=0;q--)
            printf("%d",c[q]);
        printf(" ");
        if(m!=n)
            printf(" ");
        m++;
    }
    return 0;
}