OpenJudge1001Exponentiation

问题描述

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 Rⁿwhere R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

要求

输入

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.

输出

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.

思路

本题为高精度计算问题，需要将输入的字符串转换为数组，按照四则运算的方式进行运算，具体代码如下所示：

 #include<iostream>
 #include<string>
 #include<string.h>

 using namespace std;

 struct bign{
     int data[];
     int length;
     int point;
     bign(){
         memset(data, , sizeof(data));
         length = ;
         point = ;
     }
 };

 string str;
 int n;

 bign Change(){
     bign a;
     a.point = str.length() - str.find('.') - ;
     int j = ;
     for(unsigned int i=;i<str.length();i++){
         if(str[str.length() - i - ]!='.'){
             a.data[j] = str[str.length() - i - ] - '';
             j++;
         }
     }
     a.length = j;
     return a;
 }

 bign multi(bign a, bign b){
     bign c;
     int carry = ;
     for(int i=;i<b.length;i++){
         c.length = i;
         for(int j=;j<a.length;j++){
             int temp = a.data[j] * b.data[i] + carry + c.data[i+j];
             c.data[c.length++] = temp % ;
             carry = temp / ;
         }
         while(carry != ){
             c.data[c.length++] = carry % ;
             carry /= ;
         }
     }
     c.point = a.point + b.point;
     return c;
 }

 bign power(bign a, int n){
     if(n==){
         bign d;
         d.data[] = ;
         d.length = ;
         return d;
     }
     if(n%==){
         return  multi(a, power(a, n-));
     }
     else{
         return multi(power(a, n/), power(a, n/));
     }
 }

 void show(bign rs){
     string ans;
     int zero = ;   //记录尾部0的数目
     for(int i=;i<rs.point;i++){
         if(rs.data[i]!=){
             break;
         }
         else{
             zero++;
         }
     }
     if(str[]!=''){
         for(int i=rs.length-;i>=zero;i--){
             if(zero==rs.point){
                 ans.insert(ans.end(), rs.data[i] + '');
             }
             else{
                 if(i==rs.point){
                     ans.insert(ans.end(), rs.data[i] + '');
                     ans.insert(ans.end(), '.');
                 }
                 else{
                     ans.insert(ans.end(), rs.data[i] + '');
                 }
             }
         }
     }
     else{
         ans.insert(ans.end(), '.');
         for(int i=rs.point-;i>=zero;i--){
             ans.insert(ans.end(), rs.data[i] + '');
         }
     }
     cout<<ans<<endl;
 }

 int main(){
     while(cin>>str>>n){
         bign bg = Change();
         bign rs = power(bg, n);
         show(rs);
     }
     return ;
 }

在解题过程中，遇到的是尾部零处理和首部零处理问题，要注意的是，除了题中所给的样例，10.00之类的幂次需要考虑，这类测试数据，在尾部零处理中，不应该含小数点。