hdu A+B

A+B

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

代码：

#include
#include
char a[1005],b[1005];
int c[1005];
int main()
{
    int t,m=1;
    scanf("%d",&t);
    while(m<=t)
    {
        memset(c,0,sizeof(c));
        scanf(" %s %s",a,b);
        int i,j,k,la,lb,cnt;
        la=strlen(a);
        lb=strlen(b);
        cnt=la>lb?la:lb;
        la-=1;
        lb-=1;
        for(i=0; la>=0; i++,la--)//反转大数
        {
            c[i]+=a[la]-'0';
        }
        for(i=0; lb>=0; i++,lb--)//反转并求和
        {
            c[i]+=b[lb]-'0';
            if(c[i]>=10)//处理进位
            {
                c[i]-=10;
                c[i+1]+=1;
            }
        }
        while(c[i])
        {
            if(c[i]>=10)//处理进位
            {
                c[i]-=10;
                c[i+1]+=1;
            }
            i++;
        }
        for(j=cnt; j>=0; j--)//去除前面多余的0
            if(c[j]!=0)
                break;
        printf("Case %d:\n",m++);
        printf("%s + %s = ",a,b);
        if(j!=-1)//注意答案可能为0
        {
            for(i=j; i>=0; i--)
                printf("%d",c[i]);
        }
        else
            printf("0");
        if((m-1)==t)
            printf("\n");
        else
            printf("\n\n");
    }
    return 0;
}