POJ 1001 Exponentiation 大数乘法模拟

Exponentiation

Time Limit: 500MS Memory Limit: 10000KTotal Submissions: 159520 Accepted: 38882

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

Hint

If you don't know how to determine wheather encounted the end of input: 
s is a string and n is an integer 

C++while(cin>>s>>n){...}cwhile(scanf("%s%d",s,&n)==2) //to  see if the scanf read in as many items as you want/*while(scanf(%s%d",s,&n)!=EOF) //this also work    */{...}

Source

East Central North America 1988

    题意：求取小数的n次幂，直接乘的话会丢失精度。

    分析：使用数组保存，模拟大数乘法。原理不复杂，过程略繁琐，见AC代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;const int maxn=6;char s[maxn];int  a[30*maxn],b[30*maxn],temp[30*maxn];int main(){int n;while(~scanf("%s %d",s,&n)){memset(a,0,sizeof(a));int numxsd=0;  //记录 小数点的位数int sum=0;     //记录小数部分的和  如果小数部分和为0 则说明为整数for(int i=5; i>=0; i--)if(s[i]!='.'){numxsd++;sum+=s[i]-'0';}elsebreak;if(sum==0)//整数情况{int flagd,N=s[0]-'0';//记录乘数for(int i=1; i<=5; i++)if(s[i]!='.'){N*=10;N+=s[i]-'0';}elsebreak;long long res=1;for(int i=0; i<n; i++)//直接累乘即可res*=N;printf("%lld\n",res);continue;}int num=0;//不考虑小数点  直接当作整数乘for(int i=5; i>=0; i--) //将数逆序保存在a if(s[i]!='.')a[num++]=s[i]-'0';memset(b,0,sizeof(b));for(int i=0; i<num; i++)b[i]=a[i];  //用数组b保存乘数for(int k=0; k<n-1; k++) //  每一步  a*b=temp  temp再赋值a{memset(temp,0,sizeof(temp));//每次都初始化temp数组int flag; //标记第一个不是0的位置for(int i=30*maxn-1; i>=0; i--)if(a[i]!=0){flag=i;break; //找到a[i]截止位置}for(int i=0; i<=flag; i++)for(int j=0; j<num; j++)temp[i+j]+=a[i]*b[j];//计算出tempfor(int i=0; i<30*maxn-2; i++)//对temp 进行处理if(temp[i]>=10){temp[i+1]+=temp[i]/10;temp[i]%=10;}for(int i=0; i<30*maxn; i++)//temp赋值给aa[i]=temp[i];}int flag0=0;//记录后导零位置for(int i=0; i<30*maxn; i++)  if(a[i]!=0){flag0=i;break;}int flag;for(int i=30*maxn-1; i>=0; i--)//前导0位置 if(a[i]!=0){flag=i;break;}int xsd=n*numxsd;//小数点的位置 for(int i=flag; i>=xsd; i--)printf("%d",a[i]);printf(".");//打印小数点 for(int i=xsd-1; i>=flag0; i--)printf("%d",a[i]);printf("\n");}return 0;}

     单纯的模拟大数过程的题目，大数这方面的题目需要多练习。

     特记下，以备后日回顾。