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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

代码：

#include<stdio.h>#include<string.h>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;}