大数乘法——POJ 1001

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Exponentiation

Time Limit: 500MS		Memory Limit: 10000K
Total Submissions: 141986		Accepted: 34699

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

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

题意：求实数的n次方。n<=25

思路：去掉前0后0和小数点，记录小数点位置，即为基本大数乘法。

#include<cstdio>
#include<cstdlib>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define M 10
#define L 200
char r[M];
int t[L];
char ans[L];
int n;

void Init(char *R, int &p)
{
	char tmp[M];
	memset(tmp, '\0', sizeof(tmp));
	int i, u = 0;
	int len = strlen(R);
	int end;
	for(i=0; i<len; i++) if(R[i] == '.') break;
	if(i != len)
		for(i=len-1; i>=0; i--) if(R[i] != '0') break;
	end = i;
	for(; i>=0; i--){
		if(R[i] == '.') p = i;
		else tmp[u++] = R[i]; 
	}
	int ok = 0;
	while('0' == tmp[u-1]){
		ok = 1;
		tmp[u-1] = '\0';
		u--;
	}
	for(i=0; i<u/2; i++){
		int x = tmp[i];
		tmp[i] = tmp[u-i-1];
		tmp[u-i-1] = x;
	}
	strcpy(R, tmp);
	p = end - p;
}

void Mul(char *A, const char *R)
{
	memset(t, 0, sizeof(t));
	int i, j, k = 0, u = -1;
	int lenA = strlen(A);
	int lenR = strlen(R);
	for(i=lenR-1; i>=0; i--){
		k = ++u;
		for(j=lenA-1; j>=0; j--){
			t[k++] += (R[i] - '0') * (A[j] - '0');
		}
	}
	for(i=0; i<k; i++){
		t[i+1] += t[i] / 10;
		t[i] %= 10;
	}
	if(t[k]) k++;
	for(i=0; i<k/2; i++){
		int x = t[i];
		t[i] = t[k-i-1];
		t[k-i-1] = x;
	}
	for(i=0; i<k; i++)
		A[i] = char(t[i] + '0');
}

void Pow(const char *R, int N, char *A)
{
	strcpy(A, R);
	N--;
	int i;
	for(i=0; i<N; i++)
		Mul(A, R);
}

void Print(int p, const char *A)
{
	p;
	int i;
	int len = strlen(A);
	int set_point = len - p;
	if(set_point <= 0){
		printf(".");
		for(i=0; i< -set_point; i++)
			printf("0");
	}
	for(i=0; i<len; i++){
		if(i == set_point){
			printf(".");
		}
		printf("%c", A[i]);
	}
	printf("\n");
}

int main()
{
	freopen("in.txt","r",stdin);
	while(~scanf("%s%d", r, &n))
	{
		memset(t, 0, sizeof(t));
		memset(ans, '\0', sizeof(ans));
		int p; /* 记录小数点位置 */
		Init(r, p); /* 把小数点后的尾0去掉，把第一个非零数字之前的0去掉*/
		Pow(r, n, ans); /* ans = r^n */
		Print(p*n, ans);
	}
	return 0;
}