POJ 1001 Exponentiation解题报告

Exponentiation

Time Limit: 500MS Memory Limit: 10000KTotal Submissions: 127985 Accepted: 31266

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

解题思路：解题的关键是大整数的乘法以及0和小数点的处理。

/**project:1001*version:AC+大整数乘法*author:骨灰在飞扬*Memory:236K*Time:16MS*note:*date:2014/2/10*/#include <iostream>#include <string>#include <algorithm>#include <fstream>using namespace std;const int MAX_LEN = 1000;int ia[MAX_LEN];string subF(string str1, string str2){         int i,j;     string result;     memset(ia,0,sizeof(ia));     // 暂存乘法运算的结果     reverse(str1.begin(), str1.end());     // 倒置     reverse(str2.begin(), str2.end());     int point1, point2,point;               //     小数位数     point1 = str1.find('.');         point2 = str2.find('.');     if(point1 != string::npos)               // 如果为小数，则将小数点删除          str1.erase(str1.begin()+point1);     else          point1 = 0;     if(point2 != string::npos)          str2.erase(str2.begin()+point2);     else          point2 = 0;     point1 > point2 ? point = point1 : point = point2;     // point保存最大的小数位数     int len1 = str1.size();     int len2 = str2.size();         for(i = 0; i < len1; i++)               // 乘法运算          for(j = 0; j < len2; j++)               ia[i+j] += (str1[i]-'0')*(str2[j]-'0');     int len;     // 乘积的长度     for(i = 0; i < len1+len2; i++)          // 进位处理     {          if(ia[i] >= 10)          {               ia[i+1] += (ia[i]/10);               ia[i] %= 10;          }     }     if(ia[len1+len2-1] > 0)          len = len1+len2;     else          len = len1+len2-1;     for(i = len-1; i >= 0; i--)          result += char(ia[i]+'0');     if(point > 0)          // 插入小数点     {          result.insert(result.end()-point1-point2,'.');          for(i = result.size()-1; i >= 0; i--)     // 去除小数点后的后置0          {               if(result[i] == '0')                    result.erase(result.begin()+i);               else if(result[i] == '.')   {   result.erase(result.begin()+i);   break;   }   else                    break;          }  while( (*result.begin()) == '0' )// 去除小数点前的前置0  result.erase(result.begin());     }     return result;}int main(){     //ifstream cin("E:\\vcpp\\test.txt");         string str1; int n;     while(cin >> str1 >> n)     { string str2("1"); while(n--) {           str2 = subF(str1, str2); }          cout << str2 << endl;     }     system("pause");     return 0;}