hdu1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 121378    Accepted Submission(s): 23159

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

 

 

这道题的题意很明显就是实现大数相加，所以是高精度类题目。

不过刚开始做这类题目还是要小心一些东西，例如开始输入的 时候小的小标里存放的却是大的数位，所以要反向赋值给两个整型数组（方便后面的处理），还有要注意最好不要用长度长的数字作为循环的控制条件，因为有时我们无法知道进位会进到第几位（即使知道处理起来也是比较麻烦的 ），所以是需要用最大的数字作为循环控制变量。其他的就是注意此题的输出要求比较高，要小心一些空格和回车。

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char str1[1010];
char str2[1010];
int a[1010];
int b[1010];
int c[1010];
int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
       memset(a,0,sizeof(a));
       memset(b,0,sizeof(b));
       memset(c,0,sizeof(c));
       scanf("%s",str1);
       scanf("%s",str2);
       int len1,len2;
       len1=strlen(str1);
       len2=strlen(str2);
       //输入的数字最小下标的数位最大，所以要反向赋值到a[],b[] 
       int cx=0;
       for(int j=len1-1;j>=0;j--)
       a[cx++]=str1[j]-48;
       cx=0;
       for(int j=len2-1;j>=0;j--)
       b[cx++]=str2[j]-48;
       int plus=0;
       for(int j=0;j<1002;j++)
       {
              int s;
              s=a[j]+b[j]+plus;
              c[j]=s%10;
              plus=s/10;
          
       }
       printf("Case %d:\n",i);
       printf("%s + %s = ",str1,str2);
       int j;
       for(j=1001;j>=0;j--) if(c[j]) break;
       int rec;
       rec=j;
       if(rec<0){printf("0\n");printf("\n");continue;}//以防输出的结果是0 
       for(j=rec;j>=0;j--)
       printf("%d",c[j]);
       printf("\n");
       if(i!=t)
       printf("\n");
    }
    system("pause");
    return 0;
}