hdu1002——A + B Problem II

原题：

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

 

分析：

模拟进制，字符串处理；

原码：

#include<stdio.h>

#include<string.h>

int main()

{

    char a1[1002],a2[1002];

    int c,i,j,n,m,s1[1002],s2[1002],l1,l2,count;

    count=1;

    scanf("%d",&n);

    m=n;

    while(m--)

    {

        memset(s1,0,sizeof(s1));

        memset(s2,0,sizeof(s2));//s1、s2全部赋值为0；

        scanf("%s%s",a1,a2);

        l1=strlen(a1);

        l2=strlen(a2);

        c=0;

        for(i=l1-1; i>=0; i--)

        {

            s1[c++]=a1[i]-'0';

        }

        c=0;//将字符串转换成数字

        for(i=l2-1; i>=0; i--)

        {



            s2[c++]=a2[i]-'0';

        }//将字符串转换成数字

        for(i=0; i<1002; i++)

        {

            s1[i]+=s2[i];//逐位相加

            if(s1[i]>=10)//判断是否进位

            {

                s1[i]-=10;

                s1[i+1]++;//进位

            }

        }

        printf("Case %d:\n",count++);

        printf("%s + %s = ",a1,a2);

        for(i=1001; i>=0; i--)

        {

            if(s1[i])

                break;

        }   //跳出多余的0；

        for(j=i; j>=0; j--)

        {



            printf("%d",s1[j]);//已经跳出多余的0，依次输出。

        }

        printf("\n");

        if(count!=n+1)

            printf("\n");

    }

    return 0;

}