寒假练习 03

今天刷了高精度专题，本来准备写一个高精度类，然后直接套模版，后来发现根据每题的要求分开写高精度反而效率高。

主要涉及了高精度加法、乘法、除法、取余（其中后两项为高精度和低精度进行运算）。

终于打过一遍高精度除以低精度了，高精度专题也算圆满了。

UVaOJ 424

高精度加法，水题。

#include <iostream>
#include <string>
#include <memory.h>

using namespace std;

const int MAX = 10240;

int nLen;
int pData[MAX]; 

int main()
{
	nLen = 1;
	memset(pData, 0, sizeof(pData));
	string x;
	while(cin >> x)
	{
		if(x != "0")
		{
			while(x[0] == ‘0‘) { x = x.substr(1, x.length() - 1); }
			nLen = max(nLen, (int)x.length());
			for(int i = 1; i <= x.length(); i++)
			{
				pData[i] += x[x.length() - i] - ‘0‘;
				pData[i + 1] += pData[i] / 10;
				pData[i] %= 10;
			}
			while(pData[nLen + 1]) { nLen++; }
		}
		else
		{
			for(int i = nLen; i >= 1; i--)
			{ cout << pData[i]; }
			cout << endl;
			nLen = 1;
			memset(pData, 0, sizeof(pData));
		}
	}
	return 0;
}

UVaOJ 10106

高精度乘法，要注意的是某一个乘数为0的情况，要特判一下。

#include <iostream>
#include <string>
#include <memory.h>

using namespace std;

const int MAX = 10240;

int nLen;
int pData[MAX]; 

int main()
{

	string x, y;
	while(cin >> x >> y)
	{
		if(x.length() < y.length()) { swap(x, y); }
		nLen = x.length() + y.length();
		memset(pData, 0, sizeof(pData));
		for(int i = 1; i <= y.length(); i++)
		{
			for(int j = 1; j <= x.length(); j++)
			{
				pData[i + j - 1] += (x[x.length() - j] - ‘0‘) * (y[y.length() - i] - ‘0‘);
				pData[i + j] += pData[i + j - 1] / 10;
				pData[i + j - 1] %= 10;
			}
		}
		while(nLen && !pData[nLen]) { nLen--; }
		if(nLen == 0) { nLen = 1; }
		for(int i = nLen; i >= 1; i--)
		{ cout << pData[i]; }
		cout << endl;
	}
	return 0;
}

UVaOJ 465

判断加数、乘数以及结果是否在int范围内，一开始把结果都算出来再比较。后来查了资料发现可以直接用double过，其中使用到了atof这个命令。

#include <iostream>
#include <string>
#include <stdlib.h>

using namespace std;

const int MAX_INT = 2147483647;

int main()
{
	char dwOpt;
	string x, y;
	while(cin >> x >> dwOpt >> y)
	{
		cout << x << " " << dwOpt << " " << y << endl;
		double a, b;
		a = atof(x.c_str());
		b = atof(y.c_str());
		if(a > MAX_INT) { cout << "first number too big" << endl; }
		if(b > MAX_INT) { cout << "second number too big" << endl; }
		if(dwOpt == ‘+‘ && a + b > MAX_INT) { cout << "result too big" << endl; }
		if(dwOpt == ‘*‘ && a * b > MAX_INT) { cout << "result too big" << endl; }
	}
}

UVaOJ 748

带有小数的乘方问题。

先确定小数点的位数，然后去掉小数点，进行高精度乘方运算，最后加上小数点即可。

#include <iostream>
#include <string>
#include <memory.h>

using namespace std;

const int MAX = 10240;

int nLen, nDot;
int pData[MAX]; 

string Solve(string x, string y);

int main()
{
	string x;
	int y;
	while(cin >> x >> y)
	{
		while(x[x.length() - 1] == ‘0‘) { x = x.substr(0, x.length() - 1); }
		nDot = x.length() - x.find(‘.‘) - 1;
		nDot *= y;
		x = x.substr(0, x.find(‘.‘)) + x.substr(x.find(‘.‘) + 1, x.length() - x.find(‘.‘) - 1);
		string z = "1";
		for(int i = 1; i <= y; i++)
		{ z = Solve(z, x); }
		if(z.length() <= nDot)
		{
			cout << ".";
			for(int i = z.length(); i < nDot; i++)
			{ cout << "0"; }
			cout << z << endl;
		}
		else
		{
			for(int i = 0; i < z.length() - nDot; i++)
			{ cout << z[i]; }
			cout << ".";
			for(int i = z.length() - nDot; i < z.length(); i++)
			{ cout << z[i]; }
			cout << endl;
		}
	}
	return 0;
} 

string Solve(string x, string y)
{
	if(x.length() < y.length()) { swap(x, y); }
	nLen = x.length() + y.length();
	memset(pData, 0, sizeof(pData));
	for(int i = 1; i <= y.length(); i++)
	{
		for(int j = 1; j <= x.length(); j++)
		{
			pData[i + j - 1] += (x[x.length() - j] - ‘0‘) * (y[y.length() - i] - ‘0‘);
			pData[i + j] += pData[i + j - 1] / 10;
			pData[i + j - 1] %= 10;
		}
	}
	while(nLen && !pData[nLen]) { nLen--; }
	if(nLen == 0) { nLen = 1; }
	string z = "";
	for(int i = 1; i <= nLen; i++)
	{ z = (char)(pData[i] + ‘0‘) + z; }
	return z;
}

UVaOJ 10494

高精度除以低精度以及取模运算。在网上搜了资料，终于学会了。但是要注意0 % X的情况。

另外想说的是非常喜欢这道题目的标题“If We Were A Child Again”，勾起了多少回忆。

#include <iostream>
#include <memory.h>

using namespace std;

const int MAX = 10240;

int nLen;
int pData[MAX], pAns[MAX];

string Solve(string x, long long y, char dwOpt);

int main()
{
	char dwOpt;
	string x;
	long long y;
	while(cin >> x >> dwOpt >> y)
	{ cout << Solve(x, y, dwOpt) << endl; }
	return 0;
}

string Solve(string x, long long y, char dwOpt)
{
	memset(pData, 0, sizeof(pData));
	memset(pAns, 0, sizeof(pAns));
	nLen = x.length();
	for(int i = 1; i <= x.length(); i++)
	{ pData[i] = x[x.length() - i] - ‘0‘; }
	long long d = 0;
	for(int i = nLen; i >= 1; i--)
	{
		d = d * 10 + pData[i];
		pAns[i] = d / y;
		d %= y;
	}
	while(nLen && pAns[nLen] == 0) { nLen--; }
	if(nLen == 0) { nLen = 1; }
	string z = "";
	if(dwOpt == ‘/‘)
	{
		for(int i = 1; i <= nLen; i++)
		{ z = (char)(pAns[i] + ‘0‘) + z; }
	}
	else
	{
		while(d)
		{
			z = (char)(d % 10 + ‘0‘) + z;
			d /= 10;
		}
		if(z == "") { z = "0"; }
	}
	return z;
}

今天由于题目相对比较少，所以做的比较快。接下来一个专题应该是排序和检索。