比较以科学记数法给出的两个浮点数
给定两个字符串N和M ，形式为a * 10 b 。任务是比较给定的两个浮点数并打印较小的数字，如果两个数字相等，则打印 Equal。
0<|a|<10^9 and -10^9
例子：
N and M are two numbers with two parts:
- a1 is mantissa of N and a2 is mantissa of M.
- b1 is exponent of N and b2 is exponent of M.
Input: N = 3*10^2, M = 299*10^0
Output: M
Explanation:
a1 = 3, b1 = 2
a2 = 299, b2 = 0
N = 3*10^2 = 300
M = 299*10^0 = 299.
We know that 299 is smaller than 300.
Input: N = -5*10^3, M = -50*10^2
Output : Equal
Explanation:
a1 = -5, b1 = 3
a2 = -50, b2 = 2
N = -5*10^3 = -5000
M = -50*10^2 = -5000
Hence, N and M are equal.
Input: N = -2*10^1, M = -3*10^1
Output: M
Explanation:
a1 = -2 , b1 = 1
a2 = -3 , b2 = 1
N = -20
M = -30
-30 is less than -20, hence M is smaller number.
朴素方法：我们将计算从字符串N 和 M 中提取的数字的值，然后比较哪个更好。 bigInteger 类将用于在Java中存储和计算 N 和 M 的值。对于较大的测试用例，这种方法可能会产生 Time Limit Exceeded 错误。
最佳方法：
- 从字符串N 和 M 中提取尾数和指数。
- 找到 '*' 的索引（比如说mulInd），然后 mulInd 之前的子字符串将是尾数（a1）。
- 找到 '^' 的索引，比如说powInd ，然后 powInd 之后的子字符串将是指数（b1）。
- 同样，找出 a2 和 b2。
- if(a1 > 0 && a2 < 0) ，打印M （ M 总是更小）。
- 同样，如果 (a2 > 0 && a1 < 0) ，则打印N 。
- 否则将需要使用日志进行比较。
- 以下公式将用于计算对数并得出一个结果，该结果将进行比较以确定哪个数字更大。
N = a1 * 10 ^ b1
Taking log base 10 both side
log N = log(a1) + b1 ———-(1)
Similarly, log(M) = log(a2) + b2 ———(2)
Subtract equation (1) from equation (2)
ans = (2) – (1)
ans = log(a1/a2) + b1 – b2
Therefore: int ans = log(a1/a2) + b1 – b2
- 如果 a1 < 0 ，则ans = -ans 。这是因为 a1 和 a2 都是负数。
- 如果ans < 0 ，打印N 。
- 否则，如果ans > 0 ，则打印M 。
- 否则打印相等。
下面是实现上述方法的 Jave 程序：
Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
import java.math.*;
import java.lang.*;
class GFG
{
// Function to extract mantissa
static int[] extract_mantissa(String N,
String M)
{
int mantissa[] = new int [2];
int mulInd1 = N.indexOf('*');
int a1 = Integer.parseInt(
N.substring(0, mulInd1));
mantissa[0] = a1;
int mulInd2 = M.indexOf('*');
int a2 = Integer.parseInt(
M.substring(0, mulInd2));
mantissa[1] = a2;
return mantissa;
}
// Function to extract exponent
static int[] extract_exponent(String N,
String M)
{
int exponent[] = new int [2];
int powInd1 = N.indexOf('^');
int b1 = Integer.parseInt(
N.substring(powInd1 + 1));
exponent[0] = b1;
int powInd2 = M.indexOf('^');
int b2 = Integer.parseInt(
M.substring(powInd2 + 1));
exponent[1] = b2;
return exponent;
}
// Function to find smaller number
static void solution(int a1, int b1,
int a2, int b2)
{
double x = ((double)(a1) /
(double)(a2));
double ans = (b1 - b2 +
Math.log10(x));
// If both are negative
if(a1 < 0)
ans = -ans;
if(ans < 0)
System.out.println("N");
else if(ans > 0)
System.out.println("M");
else
System.out.println("Equal");
}
static void solve(String N, String M)
{
// Extract mantissa(a1) and mantissa(a2)
// from num1 and num2
int mantissa[] = extract_mantissa(N, M);
// Extract exponent(b1) and exponent(b2)
// from num1 and num2
int exponent[] = extract_exponent(N, M);
if(mantissa[0] > 0 && mantissa[1] < 0)
System.out.println("M");
else if(mantissa[0] < 0 && mantissa[1] > 0)
System.out.println("N");
else
{
// if mantissa of both num1 and num2
// are positive or both are negative
solution(mantissa[0], exponent[0],
mantissa[1], exponent[1]);
}
}
// Driver code
public static void main (String[] args)
{
// Mantissa is negative and
// exponent is positive
String N = "-5*10^3";
String M = "-50*10^2";
solve(N, M);
// Mantissa is negative and
// exponent is negative
N = "-5*10^-3";
M = "-50*10^-2";
solve(N, M);
// Mantissa is positive and
// exponent is negative
N = "5*10^-3";
M = "50*10^-2";
solve(N, M);
// Mantissa is positive and
// exponent is positive
N = "5*10^3";
M = "50*10^2";
solve(N, M);
}
}Python3
# Python 3 program to implement
# the above approach
import math
# Function to extract mantissa
def extract_mantissa(N, M):
mantissa = [0]*2
mulInd1 = list(N).index('*')
a1 = N[0: mulInd1]
mantissa[0] = a1
mulInd2 = list(M).index('*')
a2 = M[0: mulInd2]
mantissa[1] = a2
return mantissa
# Function to extract exponent
def extract_exponent(N, M):
exponent = [0]*2
powInd1 = list(N).index('^')
b1 = N[powInd1 + 1:]
exponent[0] = b1
powInd2 = list(M).index('^')
b2 = M[powInd2 + 1:]
exponent[1] = b2
return exponent
# Function to find smaller number
def solution(a1, b1,
a2, b2):
x = int(a1) / int(a2)
ans = (int(b1) - int(b2) + math.log10(x))
# If both are negative
if(int(a1) < 0):
ans = -ans
if(ans < 0):
print("N")
elif(ans > 0):
print("M")
else:
print("Equal")
def solve(N, M):
# Extract mantissa(a1) and mantissa(a2)
# from num1 and num2
mantissa = extract_mantissa(N, M)
# Extract exponent(b1) and exponent(b2)
# from num1 and num2
exponent = extract_exponent(N, M)
if(int(mantissa[0]) > 0 and int(mantissa[1]) < 0):
print("M")
elif(int(mantissa[0]) < 0 and int(mantissa[1]) > 0):
print("N")
else:
# if mantissa of both num1 and num2
# are positive or both are negative
solution(mantissa[0], exponent[0],
mantissa[1], exponent[1])
# Driver code
if __name__ == "__main__":
# Mantissa is negative and
# exponent is positive
N = "-5*10^3"
M = "-50*10^2"
solve(N, M)
# Mantissa is negative and
# exponent is negative
N = "-5*10^-3"
M = "-50*10^-2"
solve(N, M)
# Mantissa is positive and
# exponent is negative
N = "5*10^-3"
M = "50*10^-2"
solve(N, M)
# Mantissa is positive and
# exponent is positive
N = "5*10^3"
M = "50*10^2"
solve(N, M)
# This code is contributed by ukasp.C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to extract mantissa
public static int[] extract_mantissa(String N, String M)
{
int[] mantissa = new int [2];
int mulInd1 = N.IndexOf('*');
int a1 = int.Parse(N.Substring(0, mulInd1));
mantissa[0] = a1;
int mulInd2 = M.IndexOf('*');
int a2 = int.Parse(M.Substring(0, mulInd2));
mantissa[1] = a2;
return mantissa;
}
// Function to extract exponent
public static int[] extract_exponent(String N, String M)
{
int[] exponent = new int [2];
int powInd1 = N.IndexOf('^');
int b1 = int.Parse(N.Substring(powInd1 + 1));
exponent[0] = b1;
int powInd2 = M.IndexOf('^');
int b2 = int.Parse(
M.Substring(powInd2 + 1));
exponent[1] = b2;
return exponent;
}
// Function to find smaller number
static void solution(int a1, int b1,
int a2, int b2)
{
double x = ((double)(a1) /
(double)(a2));
double ans = (b1 - b2 +
Math.Log10(x));
// If both are negative
if(a1 < 0)
ans = -ans;
if(ans < 0)
Console.WriteLine("N");
else if(ans > 0)
Console.WriteLine("M");
else
Console.WriteLine("Equal");
}
static void solve(String N, String M)
{
// Extract mantissa(a1) and mantissa(a2)
// from num1 and num2
int[] mantissa = extract_mantissa(N, M);
// Extract exponent(b1) and exponent(b2)
// from num1 and num2
int[] exponent = extract_exponent(N, M);
if(mantissa[0] > 0 && mantissa[1] < 0)
Console.WriteLine("M");
else if(mantissa[0] < 0 && mantissa[1] > 0)
Console.WriteLine("N");
else
{
// if mantissa of both num1 and num2
// are positive or both are negative
solution(mantissa[0], exponent[0],
mantissa[1], exponent[1]);
}
}
// Driver code
public static void Main ()
{
// Mantissa is negative and
// exponent is positive
String N = "-5*10^3";
String M = "-50*10^2";
solve(N, M);
// Mantissa is negative and
// exponent is negative
N = "-5*10^-3";
M = "-50*10^-2";
solve(N, M);
// Mantissa is positive and
// exponent is negative
N = "5*10^-3";
M = "50*10^-2";
solve(N, M);
// Mantissa is positive and
// exponent is positive
N = "5*10^3";
M = "50*10^2";
solve(N, M);
}
}
// This code is contributed by saurabh_jaiswal.输出：
Equal
M
N
Equal
时间复杂度： O ( 1 )
根据给定的约束 |a|最多可以是 10 个长度的字符串，如果它是负数，那么它可以是 11 个长度，同样 b 也可以是 11 个数字。字符串的最大长度可以是25（11+3+11），所以我们可以考虑提取a和b的常数时间运算。
辅助空间： O（1）