HDU 1002 A + B Problem II

最新推荐文章于 2020-08-17 22:11:42 发布

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#HDU 1002 #两超长数相加 #字符串 #ACM #A + B Problem ||

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本文详细解析了一道经典算法题——大数相加问题，通过使用数组存储超长整数，并采用翻转字符串的方式逐位计算两数之和，有效解决了传统32位整数溢出的问题。代码示例清晰地展示了如何处理大数运算，包括进位和最终结果的输出格式。

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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): 439087 Accepted Submission(s): 85433
原题链接

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

题解:题目要求求A+B的值，但是会很大，所以需要用数组存，说下我的思路，两个字符数分别存两串数，长的放在num1，短的放num2，翻转后计算，一位一位加，用一个变量（k2）来表示上一位的进位，如果进位就为1，未进位就为0，num1[i] + num2[i]+k2 的值如果转换成数值后大于等于10就进位，反之就不进位，依次计算就行了，需要注意的是中间num2的字符用完了，表示当前值的变量就从num1[i] + num2[i]+k2 变成num1[i]+k2。还有最后如果k2不为0的话那么就表示前面还有一位没有放进去需要单独放个1在前面。下面是AC代码:

#include"iostream"
#include"algorithm"
#include"string"

using namespace std;

int main()
{
	int n,j=1;
	cin >> n;
	while(n--)
	{
		string num,num1,num2,num4,num5;//新数，数1，数2，数1备份，数2备份 
		cin >> num1 >> num2;
		num4 = num1,num5=num2;
		if(num1.length() < num2.length()) //长的放在num1 
			num1 = num5,num2 = num4;
		reverse(num1.begin(),num1.end());//翻转字符串 
		reverse(num2.begin(),num2.end());
		int i,k,k2=0;//下标，当前位的值，进位值 
		for(i=0;i<num1.length();i++)
		{
			if(i >= num2.length()) k = num1[i]-48 + k2;//num2用完就只用num1 
			else k = (num1[i]-48) + (num2[i]-48) + k2;
			if(k <= 9)
			{	
				num = (char)(k+48)+num;
				k2 = 0;
			}	
			else
			{
				num = (char)(k-10+48)+num;
				k2 = 1;
			}
		}
		if(k2) num = '1'+num;//如果最后还需要进位，则单独在前面放个1 
		cout << "Case " << j << ":" << endl;
		cout << num4 << " + " << num5 << " = " << num << endl;
		j++;
		if(n-1 >= 0) cout << endl;//控制格式 
	}
	return 0;
}