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：

  1 #include<iostream>
  2 #include<string>
  3 #include<string.h>
  4 
  5 using namespace std;
  6 
  7 struct bign{
  8     int data[1000];
  9     int length;
 10     int point;
 11     bign(){
 12         memset(data, 0, sizeof(data));
 13         length = 0;
 14         point = 0;
 15     }
 16 };
 17 
 18 string str;
 19 int n;
 20 
 21 bign Change(){
 22     bign a;
 23     a.point = str.length() - str.find('.') - 1;
 24     int j = 0;
 25     for(unsigned int i=0;i<str.length();i++){
 26         if(str[str.length() - i - 1]!='.'){
 27             a.data[j] = str[str.length() - i - 1] - '0';
 28             j++;
 29         }
 30     }
 31     a.length = j;
 32     return a;
 33 }
 34 
 35 bign multi(bign a, bign b){
 36     bign c;
 37     int carry = 0;
 38     for(int i=0;i<b.length;i++){
 39         c.length = i;
 40         for(int j=0;j<a.length;j++){
 41             int temp = a.data[j] * b.data[i] + carry + c.data[i+j];
 42             c.data[c.length++] = temp % 10;
 43             carry = temp / 10;
 44         }
 45         while(carry != 0){
 46             c.data[c.length++] = carry % 10;
 47             carry /= 10;
 48         }
 49     }
 50     c.point = a.point + b.point;
 51     return c;
 52 }
 53 
 54 bign power(bign a, int n){
 55     if(n==0){
 56         bign d;
 57         d.data[0] = 1;
 58         d.length = 1;
 59         return d;
 60     }
 61     if(n%2==1){
 62         return  multi(a, power(a, n-1));
 63     }
 64     else{
 65         return multi(power(a, n/2), power(a, n/2));
 66     }
 67 }
 68 
 69 void show(bign rs){
 70     string ans;
 71     int zero = 0;   //记录尾部0的数目
 72     for(int i=0;i<rs.point;i++){
 73         if(rs.data[i]!=0){
 74             break;
 75         }
 76         else{
 77             zero++;
 78         }
 79     }
 80     if(str[0]!='0'){
 81         for(int i=rs.length-1;i>=zero;i--){
 82             if(zero==rs.point){
 83                 ans.insert(ans.end(), rs.data[i] + '0');
 84             }
 85             else{
 86                 if(i==rs.point){
 87                     ans.insert(ans.end(), rs.data[i] + '0');
 88                     ans.insert(ans.end(), '.');
 89                 }
 90                 else{
 91                     ans.insert(ans.end(), rs.data[i] + '0');
 92                 }
 93             }
 94         }
 95     }
 96     else{
 97         ans.insert(ans.end(), '.');
 98         for(int i=rs.point-1;i>=zero;i--){
 99             ans.insert(ans.end(), rs.data[i] + '0');
100         }
101     }
102     cout<<ans<<endl;
103 }
104 
105 int main(){
106     while(cin>>str>>n){
107         bign bg = Change();
108         bign rs = power(bg, n);
109         show(rs);
110     }
111     return 0;
112 }

View Code

 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。