HDOJ 1063 Exponentiation

最新推荐文章于 2014-11-11 18:28:06 发布

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本文介绍了一种计算实数R的整数次幂R^n的精确方法，其中R为0.0到99.999之间的实数，n为1到25之间的整数。通过使用字符串和数组来存储和操作大数，确保了计算结果的精确性。

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Exponentiation

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5340 Accepted Submission(s): 1447

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

Sample Output

  
   548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

Source

East Central North America 1988

Recommend

PrincetonBoy

一开始以为是一道简单题，实际上也是一道简单题，只是要注意的细节很多

001.00 2 1

000.12 2 .0144

***** 0 1

01.10 2 1.21

10 2 100 (我被这个卡住了，囧)

#include <iostream>
#include <cstring>
#include <string>
using namespace std;
const int MAX=2000;
string sa;
int a[MAX],b[MAX],c[MAX];
int ib,i,j,la;

void mul(const int a[],const int b[],int c[]);
int main(int argc, char *argv[])
{
	while (cin>>sa>>ib){
		if (!ib){
			cout<<"1\n";
			continue;
		}
		memset(a,sizeof(a),0);
		memset(b,sizeof(b),0);
		memset(c,sizeof(c),0);
		la=sa.find('.');
		while (la!=-1 && sa[sa.size()-1]=='0') sa.erase(sa.size()-1,1);
		while (sa[0]=='0') sa.erase(0,1);
		a[0]=sa.size();
		la=sa.find('.');
		if (la!=-1){
			a[0]--;
			sa.erase(la,1);
			la=a[0]-la;
		}
		if (!a[0]){
			cout<<"0\n";
			continue;
		}
		c[0]=a[0];
		for (i=1;i<=a[0];++i){
			a[i]=sa[a[0]-i]-'0';
			c[i]=sa[c[0]-i]-'0';
		}
		for (i=1;i<ib;++i){
			for (j=0;j<=c[0];++j) b[j]=c[j];
			memset(c,0,sizeof(c)); 
			mul(a,b,c);
		} 
	
		if (la!=-1){
			la*=ib;
			if (la>=c[0]){
				la-=c[0]; cout<<'.';
				for (i=0;i<la;++i) cout<<'0';
				for (i=c[0];i;--i) cout<<c[i];
			}
			else{
				for (i=c[0];i>la;--i) cout<<c[i];
				if (i) cout<<'.';
				for (i=la;i;--i) cout<<c[i];
			}
		}
		else{
			for (i=c[0];i;--i) cout<<c[i];
		}
		cout<<endl;
	}
	return 0;
}

void mul(const int a[],const int b[],int c[]){
	int i,j,len;
	for (i=1;i<=a[0];++i)
		for (j=1;j<=b[0];++j){
			c[i+j-1]+=a[i]*b[j];
			c[i+j]+=c[i+j-1]/10;
			c[i+j-1]%=10;
		}
		len=a[0]+b[0];
		while (len && !c[len]) len--;
		c[0]=len;
		if (0==c[0]) c[0]++;
	return;
}

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