有一系列数字只有数字4和7,并且数字以升序排列。系列的前几个数字是4、7、44、47、74、77、444等。给定数字N ,任务是找到该数字在给定系列中的位置。
例子:
Input: N = 4
Output: 1
Explanation:
The first number in the series is 4
Input: N = 777
Output: 14
Explanation:
The 14th number in the series is 777
方法:可以使用以下观察结果解决此问题:
- 以下是给定系列中观察到的模式。这些数字可以看成是:
""
/ \
4 7
/ \ / \
44 47 74 77
/ \ / \ / \ / \
- 如我们所见,模式的2的幂增加。因此,想法是从最低有效数字开始迭代数字的数字,并将数字的位置更新为:
- 如果当前数字= 7,则添加到该位置。
- 如果当前数字= 4,则添加到该位置。
- 完成上述操作后,打印最终位置。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the position
// of the number N
void findPosition(int n)
{
int i = 0;
// To store the position of N
int pos = 0;
// Iterate through all digit of N
while (n > 0) {
// If current digit is 7
if (n % 10 == 7) {
pos = pos + pow(2, i + 1);
}
// If current digit is 4
else {
pos = pos + pow(2, i);
}
i++;
n = n / 10;
}
// Print the final position
cout << pos;
}
// Driver Code
int main()
{
// Given number of the series
int N = 777;
// Function Call
findPosition(N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the position
// of the number N
static void findPosition(int n)
{
int i = 0;
// To store the position of N
int pos = 0;
// Iterate through all digit of N
while (n > 0)
{
// If current digit is 7
if (n % 10 == 7)
{
pos = pos + (int)Math.pow(2, i + 1);
}
// If current digit is 4
else
{
pos = pos + (int)Math.pow(2, i);
}
i++;
n = n / 10;
}
// Print the final position
System.out.print(pos);
}
// Driver Code
public static void main(String[] args)
{
// Given number of the series
int N = 777;
// Function Call
findPosition(N);
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python3 program for the above approach
# Function to find the position
# of the number N
def findPosition(n):
i = 0
# To store the position of N
pos = 0
# Iterate through all digit of N
while (n > 0):
# If current digit is 7
if (n % 10 == 7):
pos = pos + pow(2, i + 1)
# If current digit is 4
else:
pos = pos + pow(2, i)
i += 1
n = n // 10
# Print the final position
print(pos)
# Driver Code
if __name__ == '__main__':
# Given number of the series
N = 777
# Function Call
findPosition(N)
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the position
// of the number N
static void findPosition(int n)
{
int i = 0;
// To store the position of N
int pos = 0;
// Iterate through all digit of N
while (n > 0)
{
// If current digit is 7
if (n % 10 == 7)
{
pos = pos + (int)Math.Pow(2, i + 1);
}
// If current digit is 4
else
{
pos = pos + (int)Math.Pow(2, i);
}
i++;
n = n / 10;
}
// Print the final position
Console.Write(pos);
}
// Driver Code
public static void Main()
{
// Given number of the series
int N = 777;
// Function Call
findPosition(N);
}
}
// This code is contributed by Code_Mech
Javascript
输出:
14
时间复杂度: O(log 10 N)