仅打印数字 0 和 1 的数字，使其总和为 N
原文:https://www . geesforgeks . org/print-numbers-with-digits-0-and-1-only-this-sum-is-n/
给定一个数字 N，任务是找到只由 0 和 1 组成的所需数字，其和等于 N。 示例:
Input: 9
Output: 1 1 1 1 1 1 1 1 1
Only numbers smaller than or equal to 9 with
digits 0 and 1 only are 0 and 1 itself.
So to get 9, we have to add 1 - 9 times.
Input: 31
Output: 11 10 10
进场:
- 将产品 p 初始化为 1，将 m 初始化为零。
- 创建存储 0 和 1 的结果整数计数的向量。
- 循环 N 并检查 N 是否是 10 的倍数如果是，获取小数并通过乘以 10 来更新 p，并将该值存储在向量中，并将 N 减少 m。对每个小数执行此操作，并打印向量的总大小。
- 最后遍历向量并打印元素。
下面是上述方法的实现。
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the numbers
int findNumbers(int N)
{
// Initialize vector array that store
// result.
vector<int> v;
// Get the each decimal and find its
// count store in vector.
while (N) {
int n = N, m = 0, p = 1;
while (n) {
// find decimal
if (n % 10)
m += p;
n /= 10;
p *= 10;
}
v.push_back(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (int i = 0; i < v.size(); i++)
cout << " " << v[i];
return 0;
}
// Driver code
int main()
{
int N = 31;
findNumbers(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
public class GfG{
// Function to count the numbers
public static int findNumbers(int N)
{
// Initialize vector array that store
// result.
ArrayList<Integer> v = new ArrayList<Integer>();
// Get the each decimal and find its
// count store in vector.
while (N > 0) {
int n = N, m = 0, p = 1;
while (n > 0) {
// find decimal
if (n % 10 != 0)
m += p;
n /= 10;
p *= 10;
}
v.add(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (int i = 0; i < v.size(); i++)
System.out.print(" " + v.get(i));
return 0;
}
public static void main(String []args){
int N = 31;
findNumbers(N);
}
}
// This code is contributed by Rituraj Jain
Python 3
# Python 3 implementation of
# the above approach
# Function to count the numbers
def findNumbers(N) :
# Initialize vector array that
# store result.
v = [];
# Get the each decimal and find
# its count store in vector.
while (N) :
n, m, p = N, 0, 1
while (n) :
# find decimal
if (n % 10) :
m += p
n //= 10
p *= 10
v.append(m);
# Decrement N by m for
# each decimal
N -= m
# Loop for each element of vector
# And print its content.
for i in range(len(v)) :
print(v[i], end = " ")
# Driver Code
if __name__ == "__main__" :
N = 31
findNumbers(N)
# This code is contributed by Ryuga
C
// C# implementation of the above approach
using System;
using System.Collections;
class GfG
{
// Function to count the numbers
public static int findNumbers(int N)
{
// Initialize vector array that store
// result.
ArrayList v = new ArrayList();
// Get the each decimal and find its
// count store in vector.
while (N > 0)
{
int n = N, m = 0, p = 1;
while (n > 0)
{
// find decimal
if (n % 10 != 0)
m += p;
n /= 10;
p *= 10;
}
v.Add(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (int i = 0; i < v.Count; i++)
Console.Write(" " + v[i]);
return 0;
}
// Driver code
public static void Main()
{
int N = 31;
findNumbers(N);
}
}
// This code is contributed by PrinciRaj1992
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the
// above approach
// Function to count the numbers
function findNumbers($N)
{
// Initialize vector array
// that store result.
$v = array();
// Get the each decimal and find
// its count store in vector.
while ($N)
{
$n = $N;
$m = 0;
$p = 1;
while ($n)
{
// find decimal
if ($n % 10)
$m += $p;
$n /= 10;
$p *= 10;
}
array_push($v,$m);
// Decrement N by m for
// each decimal
$N -= $m;
}
// Loop for each element of vector
// And print its content.
for ($i = 0; $i < sizeof($v); $i++)
echo " " , $v[$i];
return 0;
}
// Driver code
$N = 31;
findNumbers($N);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to count the numbers
function findNumbers(N)
{
// Initialize vector array that store
// result.
let v = [];
// Get the each decimal and find its
// count store in vector.
while (N) {
let n = N, m = 0, p = 1;
while (n) {
// find decimal
if (n % 10)
m += p;
n = parseInt(n/10);
p *= 10;
}
v.push(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (let i = 0; i < v.length; i++)
document.write(" " + v[i]);
return 0;
}
// Driver code
let N = 31;
findNumbers(N);
// This code is contributed by souravmahato34.
</script>
Output:
11 10 10
时间复杂度 : O(log(N)) 辅助空间 : O(log(N))
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