HDU_oj_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：

被加数与加数太大，因此无法用普通方式计算，这时应考虑用数组存储，模拟加法运算

注意点：

①数据长度不超过1000

②每个结果后面需多空出一行

 

改了n多次

#include<iostream>
#include<cstring>
#define MAX 1000

int main()
{
    char a[MAX+1],b[MAX+1];
    int T;
    scanf("%d",&T);
    for(int w=1;w<=T;++w)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        
        scanf("%s%s",a,b);
        
        printf("Case %d:
",w);
        printf("%s + %s = ",a,b);
        
        int len1=strlen(a);
        int len2=strlen(b);
        
        int fpla=MAX-(len1>len2 ? len1:len2);
        
        for(int i=len1-1,k=MAX;i>=0;--i,--k) 
            a[k]=(a[i]-'0'),a[i]=0;
             
        for(int i=len2-1,k=MAX;i>=0;--i,--k) 
            b[k]=(b[i]-'0'),b[i]=0; 
            
        for(int i=MAX;i>=fpla;--i)
            a[i]+=b[i]; 
            
        for(int i=MAX;i>=fpla;--i)
        {
            int c=a[i]/10;
            a[i-1]+=c;
            a[i]%=10;
        } 
        for(int i=fpla+1;i<=MAX;++i)
            printf("%d",a[i]);
        puts("");
        if(w<T)
            puts("");
    }
    return 0;
}