高精度加法——杭电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

这道题是一个简单的高精度加法，对于第一次接触大数的人来说可能会有点难，一般都是因为不懂得原理。
所以只要理解了原理，就会变得简单了。

大数相加，由于已经超出了C的数值范围，所以我们只能将我们的数值用字符串处理，然后通过模拟
我们在草稿纸上打草稿。

下面是我写的代码：

#include
#include
#include

int main()
{
    char a[1000],b[1000],sum[1001];
    int la,lb,t,i,j,f,n,max,temp,carry;

    scanf("%d",&t);

    for(n=1;n<=t;n++)
    {
        carry=0;
        scanf("%s%s",a,b);
        printf("Case %d:\n",n);
        printf("%s + %s = ",a,b);

        la=strlen(a)-1;lb=strlen(b)-1; //读出两个数的长度，用于后面计算

        max=la>lb? la:lb; 

        //首先要将两个数右对齐，也就是个位对齐，然后模拟我们平时打草稿
        for(i=la,j=lb,f=max;i>=0 || j>=0;i--,j--,f--) 
        {
           temp=carry; //用于进位

           if(i>=0) temp+=a[i]-'0'; //注意：相加的时候要转换为数值
           if(j>=0) temp+=b[j]-'0';

           if(temp>9)     //如果相加大于9的话就要进位
           {
               sum[f]=temp-10+'0'; 
               //计算完后要记得转换为字符，也可以不转换，这就要看你前面是怎么定义的
               carry=1;
            }
            else 
            {
                sum[f]=temp+'0';
                carry=0;
            }

        }

        if(carry==1) printf("1"); //这里是用来进位的，就是处理像5+5=10的这类情况。
        for(i=0;i<=max;i++)  
        printf("%c",sum[i]);

       if(n==t) printf("\n");
       else printf("\n\n");

    }

    return 0;
}