A + B Problem II

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>
#define N 1001

int main(){
    int T;
    char A[N];
    char B[N];
    int sum[N];
    char temp[N];
    int A_length;
    int B_length;
    int i;
    int length;   //长度差
    int first;
    int second;
    int flag;
    int flag2;
    int time;

    scanf("%d",&T);
    time=1;    //time表示第几个输入数据组

    while(T--){
        flag=0;
        flag2=0;
        scanf("%s%s",A,B);

        printf("Case %d:
",time);
        time++;
        printf("%s + %s = ",A,B);

        A_length=strlen(A);
        B_length=strlen(B);

        if(A_length<B_length){   //保证A的长度大于或者等于B的长度
            strcpy(temp,A);
            strcpy(A,B);
            strcpy(B,temp);
        }
        
        A_length=strlen(A);
        B_length=strlen(B);
        length=A_length-B_length;
        for(i=B_length-1;i>=0;i--){  //数组B往后移动
            B[i+length]=B[i];
        }

        for(i=0;i<length;i++)   //数组B前面添加'0'
            B[i]='0';

        B[A_length]='';  //数组B添加结束符号

        for(i=A_length-1;i>=0;i--){
            first=A[i]-'0';
            second=B[i]-'0';
            sum[i]=(first+second)%10;   
            
            if(i>0)
                A[i-1]=A[i-1]-'0'+(first+second)/10+'0';
        }

        first=A[0]-'0';
        second=B[0]-'0';

        flag=(first+second)/10;

        if(flag)
            printf("%d",flag);

        for(i=0;i<A_length;i++){
            if(flag){
                flag2=1;
                printf("%d",sum[i]);
            }

            else{
                if(sum[i]!=0){
                    flag2=1;
                    printf("%d",sum[i]);
                }

                else{
                    if(flag2==1)
                        printf("%d",sum[i]);
                }
            }

        }

        if(flag2==0)
            printf("0");

        printf("
");

        if(T!=0)
            printf("
");
    }
        
    return 0;
}