HDU 1002 高精度加法

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本文解析了一个经典的编程问题“A+B Problem II”，介绍了如何通过读取两个大整数并计算它们的和，同时展示了完整的代码实现过程。

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 418680 Accepted Submission(s): 81235

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

Recommend

We have carefully selected several similar problems for you: 1004 1003 1008 1005 1089

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
	char a[1001],b[1001];
	int n,i,lena,lenb,len_max,x=1,k;
	scanf("%d",&n);
	getchar ();
	while(n--)
	{
		int a1[1001]={0},b1[1001]={0},s[2002]={0};
		scanf("%s",a);
		lena=strlen(a);
		//转为数字，逆序保存
		for(i=0;i<lena;i++)
		{
			a1[i]=a[lena-1-i]-'0';
		 } 
		 scanf("%s",b);
		 lenb=strlen(b);
		 for(i=0;i<lenb;i++)
		 {
		 	b1[i]=b[lenb-1-i]-'0';
		 }
		 if(lena>lenb)
		 len_max=lena;
		 else len_max=lenb;
		 k=0;
		 //边加边实现进位 
		 for(i=0;i<len_max;i++)
		 {
		 	s[i]=(a1[i]+b1[i]+k)%10;
		 	k=(a1[i]+b1[i]+k)/10;
		 }
		 //最高位进位情况
		 if(k!=0) s[len_max]=1;
		 printf("Case %d:\n",x);
		 x++;
		 printf("%s + %s = ",a,b);
		 if(s[len_max]==1)
		  printf("1");
		  for(i=len_max-1;i>=0;i--) //逆序输出
		  {
		  	printf("%d",s[i]);
			}
			printf("\n");
			if(n>=1) printf("\n"); 
	}return 0;
}

这个题很早以前接触过。。。很久很久之前前面的文章写到过。