【A + B Problem II 】【HDU - 1002 】（高精度加法）

原创于 2019-06-12 17:10:05 发布 · 346 阅读

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本文介绍了一种解决两个大整数相加的问题，通过模拟大数加法避免了传统整数运算的溢出问题。文章详细解释了如何读取和处理大数，以及如何进行逐位相加并处理进位的过程。

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题目：

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

解题报告：

题目意思非常简单，就是求解两个数的加和，但是由于二者非常大，会爆unsigned，所以要使用大数加法来模拟实现。

模拟起来没有什么特别的难处，有一点需要去考虑就是两个都是0的情况，或者是多个0的缀和数字。

赠送几组样例

0 0

000 00000

1000 0001

99 10

9 1

9 8

基本上注意这些，就可以正常ac 了，难在0 的这个地方，容易忽略。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;


int T;

int main()
{
	scanf("%d",&T);
	for(int kase=1;kase<=T;kase++)
	{
		char a[1010];
		char b[1010];
		int numa[1010]={0};
		int numb[1010]={0};
		int numc[1010]={0};
		scanf("%s",a);
		scanf("%s",b);
		
		int len1=strlen(a);
		int len2=strlen(b);
		int cnt1=0,cnt2=0;
		for(int i=len1-1;i>=0;i--)
		{
			numa[cnt1++]=a[i]-'0';
		}
		for(int i=len2-1;i>=0;i--)
		{
			numb[cnt2++]=b[i]-'0';
		}
		int length=max(cnt1,cnt2);
		int c=0;
		for(int i=0;i<length;i++)
		{
			c=numa[i]+numb[i]+c;
//			printf("c: %d\n",c);
			numc[i]=c%10;
			if(i!=length-1)
				c=c/10;
		}
//		printf("%d\n",c);
		if(c>=10)
		{
			numc[length]=c/10;
			numc[length-1]=c%10;
			//length++;
		}
//		printf("before: %d\n",length);
		while(numc[length]==0)
			length--; 
//		printf("after: %d\n",length);
//		printf("length: %d\n",length+1);
		printf("Case %d:\n",kase);
		printf("%s",a);
		printf(" + ");
		printf("%s",b);
		printf(" = ");
		if(length<0)
		{
			printf("0\n");
		}
		else
		{
			for(int i=length;i>=0;i--)
			{
				printf("%d",numc[i]);
			}
			printf("\n");
		}
		if(kase!=T)
				printf("\n");
	}	
}