ACM----HDU-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

 

Author

Ignatius.L

 

解题思路：根据题目要求，要实现两个非常大的整数相加，C++的数据类型已经无法满足题目要求。所以考虑按位对大数进行相加，而且特别需要注意进位问题。本位数字=（a+b+上一位的进位）%10，本位的进位=（a+b+上一位进位）/10。

注意：题目所要求的的输出格式及换行。

代码实现：

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

int main(){
    int f;
    int i,an,bn,k,n,j;
    char a[1000],b[1000];
    scanf("%d",&n);
    getchar();
    for(j=0;j<n;j++){
        int af[1001]={0},bf[1001]={0},c[1001]={0};
        scanf("%s%s",a,b);
        an=strlen(a);bn=strlen(b);
        k=an>bn?an:bn; 
        for(i=0;i<an;i++)
            af[i]=a[an-i-1]-'0';
        for(i=0;i<bn;i++)
            bf[i]=b[bn-i-1]-'0'; 
        f=0;
        for(i=0;i<k;i++){
            c[i]=(af[i]+bf[i]+f)%10;
            f=(af[i]+bf[i]+f)/10;
        }   
        printf("Case %d:\n",j+1);
        printf("%s + %s = ",a,b);
        if(f!=0)  printf("1");  
        for(i=k-1;i>=0;i--) printf("%d",c[i]);
        printf("\n");
        if(j<n-1)printf("\n");        
    }
    
}