hdu2057 A + B Again

最新推荐文章于 2024-07-21 10:00:00 发布

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#hdu2057 #A + B Again #大数加法 #大数减法 #十六进制运算

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本文介绍了一种实现两个十六进制数加法的方法，包括一个基于大数处理的复杂版本和一个简单的直接处理版本。复杂版本通过自定义数组和进位处理实现了加法逻辑，而简单版本则利用了现代编程语言的内置支持。

题意：求2个十六进制数加法的和

写了2个版本的代码

一个版本按照大数求解

#include <iostream>
#include <map>
using namespace std;

#define N 200

char str1[N];
char str2[N];
int num1[N];
int num2[N];
int num3[N];
int m_hash[256];
void init()
{
	m_hash['0'] = 0;
	m_hash['1'] = 1;
	m_hash['2'] = 2;
	m_hash['3'] = 3;
	m_hash['4'] = 4;
	m_hash['5'] = 5;
	m_hash['6'] = 6;
	m_hash['7'] = 7;
	m_hash['8'] = 8;
	m_hash['9'] = 9;
	
	m_hash['A'] = 10;
	m_hash['B'] = 11;
	m_hash['C'] = 12;
	m_hash['D'] = 13;
	m_hash['E'] = 14;
	m_hash['F'] = 15;

	m_hash['a'] = 10;
	m_hash['b'] = 11;
	m_hash['c'] = 12;
	m_hash['d'] = 13;
	m_hash['e'] = 14;
	m_hash['f'] = 15;
}

//add 方法的flag1与flag2符号必须相同，即符号相同的数相加
//flag1与flag2同时为true 或者同时为false

bool add(int num1[], int & n1, int flag1, int num2[], int n2, int flag2)
{
	int i, j, k, tmp, carry;
	bool flag = flag1;
	memset(num3, 0, sizeof(num3));

	if (n1 < n2)
	{
		memcpy(num3, num1, sizeof(int) * n1);
		memcpy(num1, num2, sizeof(int) * n2);
		memcpy(num2, num3, sizeof(int) * n1);
		tmp = n1;
		n1 = n2;
		n2 = tmp;
	}

	i = n1 - 1;
	j = n2 - 1;
	k = 0;
	carry = 0;
	while(j >= 0)
	{
		carry = num1[i] + num2[j] + carry;
		num3[k++] = carry % 16;
		carry /= 16;
		--i;
		--j;
	}

	while(i >= 0)
	{
		carry = num1[i] + carry;
		num3[k++] = carry % 16;
		carry /= 16;
		--i;
	}

	while(carry)
	{
		num3[k++] = carry % 16;
		carry /= 16;
	}
	i = k - 1;
	while(i >= 0 && num3[i] == 0) --i;
	if (i == -1)
	{
		++i;
	}

	for (j = 0; i >= 0; --i)
	{
		num1[j++] = num3[i];
	}

	n1 = j;
	if (n1 == 1 && num1[0] == 0)
	{
		flag = true;
	}
	return flag;
}

//sub 方法的flag1与flag2符号必须相异，即符号不同的数进行相减操作
//flag1与flag2必须一个为true 另一个为false
bool sub(int num1[], int & n1, bool flag1,  int num2[], int n2, bool flag2)
{
	int i, j, k, carry, tmp;
	bool flag = flag1;
	if (n1 < n2)
	{
		memcpy(num3, num1, sizeof(int) * n1);
		memcpy(num1, num2, sizeof(int) * n2);
		memcpy(num2, num3, sizeof(int) * n1);

		tmp = n1;
		n1 = n2;
		n2 = tmp;
		flag = flag2;
	}
	else if (n1 == n2)
	{
		i = 0;
		while(i < n1 && num1[i] == num2[i]) ++i;
		if (i < n1 && num1[i] < num2[i])
		{
			memcpy(num3, num1, sizeof(int) * n1);
			memcpy(num1, num2, sizeof(int) * n2);
			memcpy(num2, num3, sizeof(int) * n1);
			flag = flag2;
		}
	}
	
	i = n1 - 1;
	j = n2 - 1;
	while(j >= 0)
	{
		carry = num1[i] - num2[j];
		if(carry < 0)
		{
			num1[i - 1]--;
			num1[i] = carry + 16;
		}
		else
		{
			num1[i] = carry;
		}
		--i;
		--j;
	}

	while(i >= 0)
	{
		carry = num1[i];
		if (carry < 0)
		{
			num1[i - 1]--;
			num1[i] = carry + 16;
		}
		else
		{
			num1[i] = carry;
		}
		--i;
	}

	i = 0;
	while(i < n1 && num1[i] == 0) ++i;

	if (i == n1)
	{
		--i;
		flag = true;
	}

	for (j = 0; i < n1; ++i)
	{
		num1[j++] = num1[i];
	}
	n1 = j;
	return flag;
}

int main()
{
	int t, i,j, n1, n2, len1, len2;
	bool flag1, flag2, flag;
	init();
	while(scanf("%s %s", str1, str2) != EOF)
	{
		flag1 = false;
		flag2 = false;

		if (str1[0] != '-')
		{
			flag1 = true;;
		}

		if (str2[0] != '-')
		{
			flag2 = true;
		}

		n1 = 0, n2 = 0;
		i = 0;
		len1 = strlen(str1);
		len2 = strlen(str2);

		if (str1[i] == '+' || str1[i] == '-')
		{
			++i;
		}
		for (;i < len1; ++i)
		{
			num1[n1++] = m_hash[str1[i]];
		}

		i = 0;
		if (str2[i] == '+' || str2[i] == '-')
		{
			++i;
		}
		for (;i < len2; ++i)
		{
			num2[n2++] = m_hash[str2[i]];
		}


		if (flag1 == flag2)
		{
			flag = add(num1, n1, flag1, num2, n2, flag2);
		}
		else 
		{
			flag = sub(num1, n1, flag1, num2, n2, flag2);
		
		}
		if (!flag)
		{
			printf("-");
		}

		for (i = 0;i < n1; ++i)
		{
			printf("%c", num1[i] >= 10 ? (num1[i] - 10+ 'A'):(num1[i] + '0') );
		}
		printf("\n");
	
	}


	return 0;
}

一个简单的版本

#include <iostream>
using namespace std;

int main()
{
	__int64 a,b;
	while (scanf("%I64X %I64X",&a,&b) > 1)
	{ 
		if (a+b<0) 
		{
			printf("-%I64X\n",(-a)+(-b));
		}
		else
		{
			printf("%I64X\n",a+b);
		}
	} 
	return 0;
}