1002 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

分析

给定T个测试用例（1<=T<=20)，每个测试用例都为正数，该正数很大，不超过一千位，意味着我们不能用int、long long int，只能用字符串进行操作。

代码

//2019-7-14 13:45
//by changfengpolang
//big data A+B
#include
#include

int main()
{
    //字符串最后一位为'\0'，因此分配1001个空间 
    char a1[1001]={'\0'}, b1[1001]={'\0'}; 
    //sum保存每一位的运算结果，因为最高位可能还会进位，因此分配1001个空间  
    int a2[1000], b2[1000], sum[1001];
    
    int i=1, j, num, k, r, w;
    int len1, len2;
    //输入测试用例个数 
    scanf("%d", &num);
    while(i<=num)
    {
        //初始化 
        for(j=0;j<1000;j++)               
            a2[j]=0;
        for(j=0;j<1000;j++)
            b2[j]=0;
        for(j=0;j<1001;j++)
            sum[j]=0;
            
        scanf("%s", &a1);
        scanf("%s", &b1);
        len1 = strlen(a1);
        len2 = strlen(b1);
        
        //反转
        //或是按照自己的习惯，只要从低位到高位，每一位都对其运算即可 
        r=0;    
        for(j=len1-1;j>=0;j--)               
        {
            a2[r]=a1[j]-48;
            r++;
        }
        r=0;
        for(j=len2-1;j>=0;j--)
        {
            b2[r]=b1[j]-48;
            r++;
        }
        
        //取字符串a1和b1中较长的字符串的个数 
        if(len1=10)
            {
                sum[j]=sum[j]-10;
                w = j+1;
                sum[w]++;
            }
        }
        
        //输出结果
        printf("Case %d:\n", i);    
        for(j=0;j

重点

1.格式一定要用题目要求的格式，不然即使自己测试的结果是对的，但是格式不对，系统仍然会判错误。
2.从char类型过渡到int类型时，要倒序输入，是因为假设字符串输入“123456”时，在字符串中‘6’是最高位，但是在加法计算过程中‘6’是最低位，因此需要转换一下，方便计算。