原文:
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
using namespace std;
// function to count the numbers
int findnumbers(int n)
{
// initialize vector array that store
// result.
vector 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 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.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）
java 描述语言
output:
11 10 10
时间复杂度 : o(log(n)) 辅助空间 : o(log(n))
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