【POJ】1001 Exponentiation

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

95.123 120.4321 205.1234 156.7592  998.999 101.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201

大数，求一个数R的n次幂。

具体思路是把小数点移除，记录小数点的位置，转化成整数大数进行循环乘积，

再把小数点插入回已经乘积完成的大数数组中。

记得注意小数点末尾0的消除以及大数的进位。

怒贴代码：

#include<stdio.h>#include<iostream>#include<string.h>using namespace std;const int size=1000;  //大数位数void mult(char* A,char* B,char* ans){int i,j,k;int fract;   //总小数个数int dot=-1;  //小数点位置for(k=0;A[k]!='\0';k++)if(A[k]=='.')dot=k;int lena=k;if(dot==-1)fract=0;elsefract=lena-dot-1;dot=-1;for(k=0;B[k]!='\0';k++)if(B[k]=='.')dot=k;int lenb=k;if(dot==-1)fract+=0;elsefract+=(lenb-dot-1);  //总小数个数int a[size+1]={0};int b[size+1]={0};int pa=0,pb=0;/*倒序*/for(i=lena-1;i>=0;i--){if(A[i]=='.')continue;a[pa++]=A[i]-'0';}for(j=lenb-1;j>=0;j--){if(B[j]=='.')     //暂时删除小数点continue;b[pb++]=B[j]-'0';}int c[2*size+1]={0};int lenc;for(pb=0;pb<lenb;pb++){int w=0;  //低位到高位的进位for(pa=0;pa<=lena;pa++)  // = 为了处理最后的进位{int temp=a[pa]*b[pb]+w;w=temp/10;temp=(c[pa+pb]+=temp%10);c[lenc=pa+pb]=temp%10;w+=temp/10;}}/*倒序，得到没有小数点的ans*/for(pa=0,pb=lenc;pb>=0;pb--)ans[pa++]=c[pb]+'0';ans[pa]='\0';lena=pa; /*插入小数点*/bool flag=true; //标记是否需要删除小数末尾的0if(fract==0)   //小数位数为0，无需插入小数点flag=false;else if(fract<lena) //小数位数小于ans长度，在ans内部插入小数点{ans[lena+1]='\0';for(i=0,pa=lena;pa>0;pa--,i++){if(i==fract){ans[pa]='.';break;}elseans[pa]=ans[pa-1];}}else //小数位数大于等于ans长度，在ans前面恰当位置插入小数点{char temp[size+1];strcpy(temp,ans);ans[0]='0';ans[1]='.';for(int i=0;i<fract-lena;i++)  //补充0ans[i+2]='0';for(j=i,pa=0;pa<lena;pa++)ans[j++]=temp[pa];ans[j]='\0';}/*删除ans小数末尾的0*/if(flag){lena=strlen(ans);pa=lena-1;while(ans[pa]=='0')ans[pa--]='\0';if(ans[pa]=='.')   //小数全为0ans[pa--]='\0';}/*删除ans整数开头的0，但至少保留1个0*/pa=0;while(ans[pa]=='0')  //寻找ans开头第一个不为0的位置pa++;if(ans[pa]=='\0')  //没有小数{ans[0]='0';ans[1]='\0';}else  //有小数{for(i=0;ans[pa]!='\0';i++,pa++)ans[i]=ans[pa];ans[i]='\0';}return;}char a[size+1];char ans[size*size+1];int main(void){int b;while(cin>>a>>b){memset(ans,'\0',sizeof(ans));ans[0]='1';ans[3]='\0';for(int i=1;i<=b;i++)mult(a,ans,ans);cout<<ans<<endl;}return 0;}