【POJ】 1001-Exponentiation

标签：poj   1001   exponentiation   

题目：

Description

Input

Output

Sample Input

95.123 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

思路：

把输入全部转换成整数，剔除小数点，按整数高精度运算后再加入小数点

代码：

//21:10  --  23:11
#include <stdio.h>
#include <string.h>

char res[10006], sa[10006], sb[10006], tsb[10006];

void show(char *s)
{
	char *p;
	if (*s == 0 && *(s + 1) == -2)
		++s;
	if (*s == -2 && *(s + 1) == -1)
	{
		printf("0");
		return;
	}
	
	p = s;
	while (*++p != -1)
		;
	--p;
	while (*p != -2 && *p == 0)
		--p;
	if (*p == -2)
		--p;
	++p;
	
	while (s != p)
	{
		if (*s == -2)
			putchar('.');
		else
			putchar('0' + *s);
		++s;
	}
}
void pre(char *s)
{
	while (*s != '\0')
	{
		if (*s == '.')
			*s = -2;
		else
			*s -= '0';
		++s;
	}
	*s = -1;
}
void owncpy(char *s, char *p)
{
	while (*p != -1)
	{
		*s++ = *p++;
	}
	*s = -1;
}
char* multiple(char *s, char *a, char *b)
{
	memset(s, 0, sizeof(char) * 10006);
	int i, flags = 0, lena = 0, lenb = 0, ea, eb;
	for (i = 0; a[i] != -1; ++i)
	{
		if (flags == 1)
		{
			a[i - 1] = a[i];
			lena++;
		}
		if (a[i] == -2)
		{
			flags = 1;
			lena = 0;
		}
	}
	ea = flags ? i - 2 : i - 1;
	flags = 0;
	for (i = 0; b[i] != -1; ++i)
	{
		if (flags == 1)
		{
			b[i - 1] = b[i];
			lenb++;
		}
		if (b[i] == -2)
		{
			flags = 1;
			lenb = 0;
		}
	}
	eb = flags ? i - 2 : i - 1;

	int x, y, c;
	s[10001] = -1;
	for (x = ea; x >= 0; --x)
	{
		for (y = eb; y >= 0; --y)
		{
			c = 10000 - (ea - x + eb - y);
			s[c] += a[x] * b[y];
			if (s[c] > 9)
			{
				int tp = s[c] / 10;
				s[c] %= 10;
				s[c - 1] += tp;
			}
		}
	}

	int first = 10000 - (ea + eb + 5);
	while (first <= 10000 && s[++first] == 0)
		;
	int f = 0, ploc;
	if (lena + lenb >= 10000 - first + 1)
	{
		s[f++] = 0;
		s[f++] = -2;
		int k = lena + lenb - (10000 - first + 1);
		while (k--)
			s[f++] = 0;
	}
	else
	{
		ploc = 10000 - lena - lenb;
		while (first <= ploc)
			s[f++] = s[first++];
		s[f++] = -2;
	}
	
	while (first <= 10000)
		s[f++] = s[first++];
	s[f] = -1;
	return s;
}
void ownpow(int n)
{
	while (--n)
	{
		multiple(res, sa, sb);
		owncpy(sa, res);
		owncpy(sb, tsb);
	}
}
int main()
{
	int n;
	while (~scanf("%s %d", sa, &n))
	{
		pre(sa);
		owncpy(sb, sa);
		owncpy(tsb, sa);
		owncpy(res, sa);
		ownpow(n);
		show(res);
		printf("\n");
	}
	return 0;
}

【POJ】 1001-Exponentiation

标签：poj   1001   exponentiation   

原文地址：http://blog.csdn.net/xianyun2009/article/details/41687533