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[1000];
    int length;
    int point;
    bign(){
        memset(data, 0, sizeof(data));
        length = 0;
        point = 0;
    }
};

string str;
int n;

bign Change(){
    bign a;
    a.point = str.length() - str.find('.') - 1;
    int j = 0;
    for(unsigned int i=0;i<str.length();i++){
        if(str[str.length() - i - 1]!='.'){
            a.data[j] = str[str.length() - i - 1] - '0';
            j++;
        }
    }
    a.length = j;
    return a;
}

bign multi(bign a, bign b){
    bign c;
    int carry = 0;
    for(int i=0;i<b.length;i++){
        c.length = i;
        for(int j=0;j<a.length;j++){
            int temp = a.data[j] * b.data[i] + carry + c.data[i+j];
            c.data[c.length++] = temp % 10;
            carry = temp / 10;
        }
        while(carry != 0){
            c.data[c.length++] = carry % 10;
            carry /= 10;
        }
    }
    c.point = a.point + b.point;
    return c;
}

bign power(bign a, int n){
    if(n==0){
        bign d;
        d.data[0] = 1;
        d.length = 1;
        return d;
    }
    if(n%2==1){
        return  multi(a, power(a, n-1));
    }
    else{
        return multi(power(a, n/2), power(a, n/2));
    }
}

void show(bign rs){
    string ans;
    int zero = 0;   //记录尾部0的数目
    for(int i=0;i<rs.point;i++){
        if(rs.data[i]!=0){
            break;
        }
        else{
            zero++;
        }
    }
    if(str[0]!='0'){
        for(int i=rs.length-1;i>=zero;i--){
            if(zero==rs.point){
                ans.insert(ans.end(), rs.data[i] + '0');
            }
            else{
                if(i==rs.point){
                    ans.insert(ans.end(), rs.data[i] + '0');
                    ans.insert(ans.end(), '.');
                }
                else{
                    ans.insert(ans.end(), rs.data[i] + '0');
                }
            }
        }
    }
    else{
        ans.insert(ans.end(), '.');
        for(int i=rs.point-1;i>=zero;i--){
            ans.insert(ans.end(), rs.data[i] + '0');
        }
    }
    cout<<ans<<endl;
}

int main(){
    while(cin>>str>>n){
        bign bg = Change();
        bign rs = power(bg, n);
        show(rs);
    }
    return 0;
}

 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。