A + B Problem II

最新推荐文章于 2023-06-04 16:18:52 发布

sping-

最新推荐文章于 2023-06-04 16:18:52 发布

阅读量235

点赞数

CC 4.0 BY-SA版权

分类专栏：刷题之旅

本文链接：https://blog.youkuaiyun.com/hpuspring/article/details/77692925

刷题之旅专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一个关于处理非常大的整数相加的问题，并提供了一种通过字符串操作来实现大数加法的方法。该方法避免了使用32位整数进行处理时可能遇到的溢出问题，展示了如何读取两个大数、如何逐位相加并处理进位的过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

点击打开链接

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

本题就是考察字符串的运用，注意输出格式；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
char s1[1000005];
char s2[1000005];
int a[1000005];
int b[1000005];
int main()
{
	int T;
	scanf("%d",&T);
	int Case =0;
	while(T--)
	{    int l1,l2;
	     int k=0;
		memset(a,0,sizeof(a));
    	memset(b,0,sizeof(b));
		scanf("%s%s",s1,s2);
		l1=strlen(s1);
		l2=strlen(s2);
		int i,j;
		for( i=l1-1,j=0;i>=0;i--,j++)
		{
			 a[j]=s1[i]-'0';
		}
		for(i=l2-1,j=0;i>=0;i--,j++)
		{
			b[j]=s2[i]-'0';
		}
			if(l1<l2)
		l1=l2;
		for( i=0;i<l1;i++)
		{
			a[i]+=b[i];
			if(a[i]>=10)
			{
			a[i]-=10;
	 		a[i+1]++;
	      	}
		}
		int  n;
		if(a[i-1]>=10)
		n=i;
		else n=i-1;
		printf("Case %d:\n%s + %s = ",++Case,s1,s2);
     	for( i=n;i>=0;i--)
		printf("%d",a[i]);
		if(T!=0) 
		printf("\n\n");
		else printf("\n");
	}
	return 0;
}