杭电1063 hdu1063Exponentiation

Exponentiation

Time Limit: 2000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8385    Accepted Submission(s): 2393

Problem 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
  
  

 

Source

East Central North America 1988

 

求输入数据的N次方，高精度，调试了一个小时，原以为写那么乱AC不过呢，结果还是过了，附代码+注释：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[630],b[630],c[660],i,j,k,l,m,n,cnt;
char s[10];
void pplus(int a[])
{
	int i,l;
	memset(c,sizeof(c));
	for(i=0;i<10;i++)//大数乘模板 
	for(j=0;j<200;j++)
	c[i+j]+=b[j]*a[i];
	
	for(i=0;i<200;i++)
	{
		if(c[i]>=10)
		c[i+1]+=c[i]/10;
		c[i]%=10;
	}
	for(i=0;i<300;i++)
	b[i]=c[i];
}
int main()
{
	while(scanf("%s%d",s,&n)!=EOF)
	{
		l=strlen(s);
		cnt=0; 
		for(i=l-1;i>=0;i--)//判断小数点后有几位数 
		if(s[i]=='.')
		cnt=l-1-i;
		int help=0;
		memset(c,sizeof(c));
		memset(b,sizeof(b));
		int nmd=0;
		
		for(i=l-1;i>=0;i--)//先暂时消除输入数据的后置0 
		if(s[i]!='0')
		break;
		
		if(s[i]=='.')//如果输入的类似1.   这样的数据 
		nmd=1;
		for(i=l-1;i>=0;i--)//判断输入有无 小数点 
		if(s[i]=='.')
		break;
		if(i==-1)
		nmd=1;
		
		b[0]=1;
		
		
		for(i=l-1;i>=0;i--) //储存数据 
		if(s[i]>='0'&&s[i]<='9')
		a[help++]=s[i]-'0';
		
		
		for(i=0;i<n;i++) //递归相乘 
		pplus(a);
		
		
		for(i=200;i>=cnt*n;i--)//输出 
		if(b[i])
		break;
		for(i;i>=cnt*n;i--)
		printf("%d",b[i]);
		if(nmd==0) //如果结果是小数 
		printf(".");
		int cnm=0;
		for(;i>=0;i--)
		{
			cnm=0;
			for(j=i;j>=0;j--)//消除后置0，其实可以直接输入成浮点型判断的 
			{
				if(b[j])//但是我比较懒 
				{
					cnm=1;
					break;
				}
			}
			if(cnm)
			printf("%d",b[i]);
		}
		printf("\n");
	}
}