(解题报告)HDU1002---大数加法

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 278427 Accepted Submission(s): 53689

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

解题思路：
1. 测试用例用for循环控制；
2. 数据的读入使用字符串，将字符串按位储存进数组，计算时使用数字按位逐项相加（注意把两个数组倒过来，这样控制位数不出错），在判断最大位是否进1，如果进1，则使最长的数组长度加1，最后再逆序输出相加之和；
3. 使用的头文件string.h用于计算数组长度并比较出最大的一个，作为和数组的长度，再判断是否已溢出；
4. 使用 memset(数组名，0，sizeof(数组名)) 对数组进行清0；memset详细用法见[http://blog.sina.com.cn/s/blog_715d0ae30100yj2d.html ]
5. 数字的ASCLL码值减去‘0’等于数字本身；

代码如下:

#include <stdio.h>
#include <string.h>
int max(int a,int b)
{
   return  a>b?a:b;                     //判断长度大小，以最长的数组为加法的最大值 
}
 int main()
 {
   char a[1005],b[1005];                //用字符串输入数字，再用数组保存 
   int  a1[1005],a2[1005],a3[1005];
   int l1,l2,l;
   int i,n,m,k,ok=0;                   
   scanf("%d",&n);
   for (k=1;k<=n;k++)
   {
       memset(a1,0,sizeof(a1));
       memset(a2,0,sizeof(a2));
       memset(a3,0,sizeof(a3));
       scanf("%s%s",a,b);
       l1=strlen(a);
       l2=strlen(b);
       for (i=0;i<l1;i++)
        a1[i]=a[l1-1-i]-'0';         //把字符串倒着放进数组，以便于之后相加 ，最后一位数是长度减1 
       for (i=0;i<l2;i++)            //字符数字减‘0 ’得到的是数字本身 
        a2[i]=b[l2-1-i]-'0';
        l=max(l1,l2);
        for (i=0;i<l;i++)           
        {
        a3[i]=a1[i]+a2[i]+a3[i];
        if (a3[i]>=10)
        {
        a3[i]=a3[i]%10;
        a3[i+1]++;
        }
        }
        if (a3[l]==1)//如果位数进了一个，则使数组长度加1 
        l++;
        if (!ok)
            ok++;
        else
        {
            printf("\n");
        }
            printf("Case %d:\n",k);
            printf("%s + %s = ",a,b);
            for (i=l-1;i>=0;i--)
            printf("%d",a3[i]);      //再将得到的数组逆序输出 
            printf("\n");

   }
        return 0;
   }

c++代码如下：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    char a[1005],b[1005],c[1005];
    int  a1[1005],b1[1005],c1[1005];
    int l1,l2,l3,i,n,j=0;
    cin>>n;
    for(j=1;j<=n;j++)
    {
        memset(a1,0,sizeof(a1));
        memset(b1,0,sizeof(b1));
        memset(c1,0,sizeof(c1));
        cin>>a>>b;
        if(strlen(a)>=strlen(b))
        {
            l1=strlen(a);
            l2=strlen(b);
        }
        else 
        {
            l1=strlen(b);
            l2=strlen(a);
        }
        for(i=0;i<strlen(a);i++)    
        {
            a1[i]=a[strlen(a)-1-i]-'0'; 
        }
        for(i=0;i<strlen(b);i++)
        {
            b1[i]=b[strlen(b)-1-i]-'0'; 
        }

        for(i=0;i<l1;i++)
        {   
            c1[i]=a1[i]+b1[i]+c1[i];
            if(c1[i]>=10)
            {
                c1[i]%=10;
                c1[i+1]++;
            }
        }
        if(c1[l1]==1)
        {
            l1++;   
        }
        cout<<"Case "<<j<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        for(i=l1-1;i>=0;i--)
        {
            cout<<c1[i];
        }
        if(j==n)
        cout<<endl;
        else
        cout<<endl<<endl;
    }
    return 0;
}

仅代表个人看法，其他方法很多。