PAT甲级——A1060 Are They Equal

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 1, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line YES if the two numbers are treated equal, and then the number in the standard form 0.d[1]...d[N]*10^k (d[1]>0 unless the number is 0); or NO if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

其实这道题的难度在于得到这个数的幂次

#include <iostream>
#include <string>
using namespace std;
string A, B;//注意，10^100超出double的范围，只能使用字符串来存取
int N;
int dealString(string &str)//数据预处理,返回数据位数
{
    int k = str.find('.');//找到小数点
    if (k != -1)
    {
        str.erase(k, 1);//删除小数点
        if (str[0] == '0')
        {
            k = 0;
            str.erase(0, 1);//删除第一个0
        }
    }
    else//没有小数
    {
        if (str != "0")
            k = str.length();
        else
        {
            k = 0;
            str.erase(0, 1);//删除第一个0
        }
    }
    while (!str.empty() && str[0] == '0')
    {
        str.erase(0, 1);//输出前面的0
        k--;//比如0.000128 = 0.128*10^-3
    }
    if (str.empty())//这个数就是0
        k = 0;
    while (str.length() < N)
        str += "0";//位数不够0来凑
    return k;
}

int main()
{
    cin >> N >> A >> B;
    //使用k1,k2来得到A,B的位数
    int k1, k2;
    k1 = dealString(A);
    k2 = dealString(B);
    A.assign(A.begin(), A.begin() + N);//取前N位
    B.assign(B.begin(), B.begin() + N);
    if (A == B && k1 == k2)
    {
        cout << "YES ";
        cout << "0." << A << "*10^" << k1 << endl;
    }
    else
    {
        cout << "NO ";
        cout << "0." << A << "*10^" << k1 << " ";
        cout << "0." << B << "*10^" << k2 << endl;
    }
    return 0;    
}