POJ 1001Exponentiation 实数的幂

http://poj.org/problem?id=1001

WA了三次

整数的幂忘记去掉小数点了！！！！！

 

Exponentiation

Time Limit: 500MS

Memory Limit: 10000K

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 R ⁿ 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

/* Author : yan
 * Question : POJ 1001 Exponentiation
 * Date && Time : Monday, January 31 2011 11:20 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
#define MAX 500
#define bool _Bool
#define true 1
#define false 0

char ans[2*MAX];
char stack[2*MAX];
int top;

void multiply(char* a,char* b,char* c)
{
    int i,j,ca,cb,* s;
    ca=strlen(a);
    cb=strlen(b);
    s=(int*)malloc(sizeof(int)*(ca+cb));
    for (i=0;i<ca+cb;i++)
        s[i]=0;
    for (i=0;i<ca;i++)
        for (j=0;j<cb;j++)
            s[i+j+1]+=(a[i]-'0')*(b[j]-'0');//i+j+1的目的就是为了防止最高位进位而产生错误
    for (i=ca+cb-1;i>=0;i--)
        if (s[i]>=10)
        {
            s[i-1]+=s[i]/10;
            s[i]%=10;
        }
    i=0;
    while (s[i]==0)
        i++;//去除前导0
       for (j=0;i<ca+cb;i++,j++)
           c[j]=s[i]+'0';
    c[j]='/0';//将结果存储到字符数组
    free(s);
}
int del_point(char *ch)
{
	int len,i,res;
	len=strlen(ch);
	for(i=0;i<len;i++)
		if(ch[i]=='.')
		{
			res=len-i-1;
			break;
		}
	for(;i<len-1;i++) ch[i]=ch[i+1];
	ch[len-1]='/0';
	return res;
}
void add_point(char *ch,int pos)
{
	int i,len;
	bool before=false;
	len=strlen(ch);
	if(len>pos) before=true;
	top=0;
	pos++;
	while(pos--)
		if(len<0) stack[top++]='0';
		else stack[top++]=ch[len--];
	stack[top++]='.';
	while(len>0)
	{
		stack[top++]=ch[len--];
	}
	if(before)
		stack[top++]=ch[0];
}
void print_stack()
{
	int i=0;
	while(top--)
		ans[i++]=stack[top];
	ans[i]='/0';
}
void del_zero()
{
	int len;
	len=strlen(ans);
	while(len--)
	{
		if(ans[len]=='0') ans[len]='/0';
		else break;
	}
	len=strlen(ans);
	if(ans[len-1]=='.') ans[len-1]='/0';
}
int main()
{
	//freopen("input","r",stdin);
	//freopen("output","w",stdout);
	char base[7];
	int index;
	int pos,i;
	while(scanf("%s %d",base,&index)!=EOF)
	{
		pos=del_point(base);
		strcpy(ans,"1");
		for(i=0;i<index;i++)
		{
			multiply(ans,base,ans);
		}
		add_point(ans,pos*index);
		print_stack();
		//printf("%s/n",res);
		//printf("%s %d/n",ans,pos*index);
		del_zero();
		printf("%s/n",ans);
	}
	return 0;
}