大数乘法——POJ 1001

对应POJ题目：点击打开链接

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

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,'',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] = '';
		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,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,sizeof(t));
		memset(ans,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;
}