poj 1001 高精度乘法

Description

Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rⁿ where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.

Sample Input

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

#include<iostream>
#include<string.h>
#include<cstdio>
using namespace std;

#define  size 1000   //大数位数

void mult(char * A, char * B, char * ans);

char a[size + 1];
char ans[size * size + 1];

int main(void)
{
	int b;

	while (cin >> a >> b)
	{
		memset(ans, ' ', sizeof(ans));
		ans[0] = '1';
		ans[1] = ' ';

		for (int i = 1; i <= b; i++)
		{
			mult(a, ans, ans);
		}

		cout << ans << endl;
	}

	return 0;
}

void mult(char * A, char * B, char * ans)
{		//printf("
A %s
B %s
ans %s
",A,B,ans);
	int i, j, k;

	int fract;   //总小数个数
	int dot = -1; //小数点位置

	for (k = 0; A[k] != ' '; k++)
	{
		if (A[k] == '.')
		{
			dot = k;
		}
	}		//printf("a的小数点位置   %d
",dot);

	int lena = k;

	if (dot == -1)
	{
		fract = 0;
	}
	else
	{
		fract = lena - dot - 1;
	}		//printf("小数的位数   %d
",fract);

	dot = -1;

	for (k = 0; B[k] != ' '; k++)
	{
		if (B[k] == '.')
		{
			dot = k;
		}
	}		//printf("B的小数点位置 %d
",dot);

	int lenb = k;

	if (dot == -1)
	{
		fract += 0;
	}
	else
	{
		fract += (lenb - dot - 1);    //总小数个数
	}		//printf("a+b的总小数个数   %d
",fract);

	int a[size + 1] = {0};
	int b[size + 1] = {0};
	int pa = 0, pb = 0;

	/*倒序    将字符串数组变成整数数组*/

	for (i = lena - 1; i >= 0; i--)
	{
		if (A[i] == '.') //暂时删除小数点
		{
			continue;
		}

		a[pa++] = A[i] - '0';		//printf("%d",a[pa-1]);
	}		//printf("	A
");

	for (j = lenb - 1; j >= 0; j--)
	{
		if (B[j] == '.')  //暂时删除小数点
		{
			continue;
		}

		b[pb++] = B[j] - '0';		//printf("%d",b[pb-1]);
	}		//printf("	B
");

	int c[2 * size + 1] = {0};
	int lenc;

	for (pb = 0; pb < lenb; pb++)
	{
		int w = 0; //低位到高位的进位

		for (pa = 0; pa <= lena; pa++) // = 为了处理最后的进位
		{
			int temp = a[pa] * b[pb] + w;
			w = temp / 10;
			temp = (c[pa + pb] += temp % 10);
			c[lenc = pa + pb] = temp % 10;
			w += temp / 10;
		}
	}

	/*倒序，得到没有小数点的ans*/

	for (pa = 0, pb = lenc; pb >= 0; pb--)
	{
		ans[pa++] = c[pb] + '0';
	}		//puts(ans);

	ans[pa] = ' ';
	lena = pa;

	/*插入小数点*/

	bool flag = true; //标记是否需要删除小数末尾的0

	if (fract == 0) //小数位数为0，无需插入小数点
	{
		flag = false;
	}
	else
		if (fract < lena) //小数位数小于ans长度，在ans内部插入小数点
		{
			ans[lena + 1] = ' ';

			for (i = 0, pa = lena; pa > 0; pa--, i++)
			{
				if (i == fract)
				{
					ans[pa] = '.';
					break;
				}
				else
				{
					ans[pa] = ans[pa - 1];
				}
			}

		}
		else //小数位数大于等于ans长度，在ans前面恰当位置插入小数点
		{
			char temp[size + 1];
			strcpy(temp, ans);
			ans[0] = '0';
			ans[1] = '.';

			for (int i = 0; i < fract - lena; i++) //补充0
			{
				ans[i + 2] = '0';
			}

			for (j = i, pa = 0; pa < lena; pa++)
			{
				ans[j++] = temp[pa];
			}

			ans[j] = ' ';
		}

	/*删除ans小数末尾的0*/

	if (flag)
	{
		lena = strlen(ans);
		pa = lena - 1;

		while (ans[pa] == '0')
		{
			ans[pa--] = ' ';
		}

		if (ans[pa] == '.') //小数全为0
		{
			ans[pa--] = ' ';
		}
	}

	/*删除ans整数开头的0，但至少保留1个0*/

	pa = 0;

	while (ans[pa] == '0') //寻找ans开头第一个不为0的位置
	{
		pa++;
	}

	if (ans[pa] == ' ') //没有小数
	{
		ans[0] = '0';
		ans[1] = ' ';
	}
	else  //有小数
	{
		for (i = 0; ans[pa] != ' '; i++, pa++)
		{
			ans[i] = ans[pa];
		}

		ans[i] = ' ';
	}

	return;
}

poj 1001 高精度乘法

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