PAT-A 1060 Are They Equal （25 分)

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 0.123×10​5​​ 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 (<100) 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 10​100​​, 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

思路：

关键是找到小数点移动的方向和距离。规律如下

浮点数      小数点下标       第一个非0数的下标       指数
12345         5                          0                               5 (5-0)
123.45        3                          0                               3 (3-0)
0.00123·     1                          4                               -2 (1 - 4 + 1)
0.000345    1                          5                               -3  (1 - 5 + 1)

 

坑点：0要写成0.000*10^0的形式。
  

/*
浮点数      小数点下标      第一个非0数的下标       指数
12345       5               0                       5 (5-0)
123.45      3               0                       3 (3-0)
0.00123·    1               4                       -2 (1 - 4 + 1)
0.000345    1               5                       -3  (1 - 5 + 1)
*/

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>

using namespace std;

// 返回第一个小数点的位置
int indexOfPoint(char num[])
{
    for(int i = 0; i < strlen(num); i++)
    {
        if(num[i] == '.')
            return i;
    }
    return strlen(num);
}

//返回第一个非零整数的位置
int indexOfFirst(char num[])
{
    for(int i = 0; i < strlen(num); i++)
    {
        if(num[i] > '0' && num[i] <= '9')
            return i;
    }
    return -1; //返回-1表示是0
}

// 得到标准格式
string get(char n1[], int n)
{
    int point; // 小数点的下标
    int first; // 第一个非零数字的下标

    //先计算第一个浮点数
    point = indexOfPoint(n1); //小数点的位置
    first = indexOfFirst(n1); //第一个非零数字的位置

    string s1 = "0.";
    if(first != -1)
    {
        // 标准格式的指数
        int index = point - first;
        index = index > 0 ? index : index + 1;

        // 计算有效位上的数字
        int i = 0;
        while(i < n)
        {
            if(first < strlen(n1) && n1[first] != '.')
            {
                s1 += n1[first];
                i++;
            }
            else if(first >= strlen(n1))
            {
                s1 += "0";
                i++;
            }
            first++;
        }
        s1 += "*10^";
        s1 += to_string(index);
    }
    else
    {
        for(int i = 0; i < n; i++)
            s1 += "0";
        s1 += "*10^";
        s1 += "0"; // 0.000要求为0.000*10^0
    }
    return s1;
}
int main()
{
    char n1[110], n2[110];
    int n;

    scanf("%d", &n);
    scanf("%s", n1);
    scanf("%s", n2);

    string s1 = get(n1, n);
    string s2 = get(n2, n);
    if(s1 == s2)
        printf("YES %s\n", s1.c_str());
    else
        printf("NO %s %s\n", s1.c_str(), s2.c_str());
    return 0;
}