求数组中零、正数、负数的比例
给定一个由大小为 N 的整数组成的数组 a ，任务是找出数组中的正数、负数和零的比例，最多到小数点后四位。
示例:
输入: a[] = {2，-1，5，6，0，-3} 输出: 0.5000 0.3333 0.1667 有 3 个正，2 个负，1 个零。它们的比率将是正的:3/6 = 0.5000，负的:2/6 = 0.3333，零的:1/6 = 0.1667。
输入: a[] = {4，0，-2，-9，-7，1} 输出: 0.3333 0.5000 0.1667 有 2 正，3 负，1 零。它们的比率将是正的:2/6 = 0.3333，负的:3/6 = 0.5000，零的:1/6 = 0.1667。
进场:
- 计算数组中正元素的总数。
- 计算数组中负元素的总数。
- 计算数组中零元素的总数。
- 将正元素、负元素和零元素的总数除以数组的大小，得到比值。
- 打印数组中的正、负、零元素的比率，最多四位小数。
下面是上述方法的实现。
C++
// C++ program to find the ratio of positive,
// negative, and zero elements in the array.
#include<bits/stdc++.h>
using namespace std;
// Function to find the ratio of
// positive, negative, and zero elements
void positiveNegativeZero(int arr[], int len)
{
// Initialize the postiveCount, negativeCount, and
// zeroCountby 0 which will count the total number
// of positive, negative and zero elements
float positiveCount = 0;
float negativeCount = 0;
float zeroCount = 0;
// Traverse the array and count the total number of
// positive, negative, and zero elements.
for (int i = 0; i < len; i++) {
if (arr[i] > 0) {
positiveCount++;
}
else if (arr[i] < 0) {
negativeCount++;
}
else if (arr[i] == 0) {
zeroCount++;
}
}
// Print the ratio of positive,
// negative, and zero elements
// in the array up to four decimal places.
cout << fixed << setprecision(4) << (positiveCount / len)<<" ";
cout << fixed << setprecision(4) << (negativeCount / len)<<" ";
cout << fixed << setprecision(4) << (zeroCount / len);
cout << endl;
}
// Driver Code.
int main()
{
// Test Case 1:
int a1[] = { 2, -1, 5, 6, 0, -3 };
int len=sizeof(a1)/sizeof(a1[0]);
positiveNegativeZero(a1,len);
// Test Case 2:
int a2[] = { 4, 0, -2, -9, -7, 1 };
len=sizeof(a2)/sizeof(a2[0]);
positiveNegativeZero(a2,len);
}
// This code is contributed by chitranayal
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the ratio of positive,
// negative, and zero elements in the array.
class GFG {
// Function to find the ratio of
// positive, negative, and zero elements
static void positiveNegativeZero(int[] arr)
{
// Store the array length into the variable len.
int len = arr.length;
// Initialize the postiveCount, negativeCount, and
// zeroCountby 0 which will count the total number
// of positive, negative and zero elements
float positiveCount = 0;
float negativeCount = 0;
float zeroCount = 0;
// Traverse the array and count the total number of
// positive, negative, and zero elements.
for (int i = 0; i < len; i++) {
if (arr[i] > 0) {
positiveCount++;
}
else if (arr[i] < 0) {
negativeCount++;
}
else if (arr[i] == 0) {
zeroCount++;
}
}
// Print the ratio of positive,
// negative, and zero elements
// in the array up to four decimal places.
System.out.printf("%1.4f ", positiveCount / len);
System.out.printf("%1.4f ", negativeCount / len);
System.out.printf("%1.4f ", zeroCount / len);
System.out.println();
}
// Driver Code.
public static void main(String args[])
{
// Test Case 1:
int[] a1 = { 2, -1, 5, 6, 0, -3 };
positiveNegativeZero(a1);
// Test Case 2:
int[] a2 = { 4, 0, -2, -9, -7, 1 };
positiveNegativeZero(a2);
}
}
Python 3
# Python3 program to find the ratio of positive,
# negative, and zero elements in the array.
# Function to find the ratio of
# positive, negative, and zero elements
def positiveNegativeZero(arr):
# Store the array length into the variable len.
length = len(arr);
# Initialize the postiveCount, negativeCount, and
# zeroCountby 0 which will count the total number
# of positive, negative and zero elements
positiveCount = 0;
negativeCount = 0;
zeroCount = 0;
# Traverse the array and count the total number of
# positive, negative, and zero elements.
for i in range(length):
if (arr[i] > 0):
positiveCount += 1;
elif(arr[i] < 0):
negativeCount += 1;
elif(arr[i] == 0):
zeroCount += 1;
# Print the ratio of positive,
# negative, and zero elements
# in the array up to four decimal places.
print("{0:.4f}".format((positiveCount / length)), end=" ");
print("%1.4f "%(negativeCount / length), end=" ");
print("%1.4f "%(zeroCount / length), end=" ");
print();
# Driver Code.
if __name__ == '__main__':
# Test Case 1:
a1 = [ 2, -1, 5, 6, 0, -3 ];
positiveNegativeZero(a1);
# Test Case 2:
a2 = [ 4, 0, -2, -9, -7, 1 ];
positiveNegativeZero(a2);
# This code is contributed by Rajput-Ji
C
// C# program to find the ratio of positive,
// negative, and zero elements in the array.
using System;
class GFG {
// Function to find the ratio of
// positive, negative, and zero elements
static void positiveNegativeZero(int[] arr)
{
// Store the array length into the variable len.
int len = arr.Length;
// Initialize the postiveCount, negativeCount, and
// zeroCountby 0 which will count the total number
// of positive, negative and zero elements
float positiveCount = 0;
float negativeCount = 0;
float zeroCount = 0;
// Traverse the array and count the total number of
// positive, negative, and zero elements.
for (int i = 0; i < len; i++) {
if (arr[i] > 0) {
positiveCount++;
}
else if (arr[i] < 0) {
negativeCount++;
}
else if (arr[i] == 0) {
zeroCount++;
}
}
// Print the ratio of positive,
// negative, and zero elements
// in the array up to four decimal places.
Console.Write("{0:F4} ", positiveCount / len);
Console.Write("{0:F4} ", negativeCount / len);
Console.Write("{0:F4} ", zeroCount / len);
Console.WriteLine();
}
// Driver Code.
public static void Main(String []args)
{
// Test Case 1:
int[] a1 = { 2, -1, 5, 6, 0, -3 };
positiveNegativeZero(a1);
// Test Case 2:
int[] a2 = { 4, 0, -2, -9, -7, 1 };
positiveNegativeZero(a2);
}
}
// This code is contributed by sapnasingh4991
java 描述语言
<script>
// Javascript program to find the
// ratio of positive, negative, and
// zero elements in the array.
// Function to find the ratio of
// positive, negative, and zero elements
function positiveNegativeZero(arr)
{
// Store the array length into
// the variable len.
let len = arr.length;
// Initialize the postiveCount,
// negativeCount, and zeroCountby
// 0 which will count the total number
// of positive, negative and zero elements
let positiveCount = 0;
let negativeCount = 0;
let zeroCount = 0;
// Traverse the array and count the
// total number of positive, negative,
// and zero elements.
for(let i = 0; i < len; i++)
{
if (arr[i] > 0)
{
positiveCount++;
}
else if (arr[i] < 0)
{
negativeCount++;
}
else if (arr[i] == 0)
{
zeroCount++;
}
}
// Print the ratio of positive,
// negative, and zero elements
// in the array up to four decimal places.
document.write((positiveCount / len).toFixed(4) + " ");
document.write((negativeCount / len).toFixed(4) + " ");
document.write((zeroCount / len).toFixed(4));
document.write("<br>");
}
// Driver Code.
// Test Case 1:
let a1 = [ 2, -1, 5, 6, 0, -3 ];
positiveNegativeZero(a1);
// Test Case 2:
let a2 = [ 4, 0, -2, -9, -7, 1 ];
positiveNegativeZero(a2);
// This code is contributed by sravan kumar G
</script>
Output:
0.5000 0.3333 0.1667
0.3333 0.5000 0.1667
时间复杂度: O(len)
辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
