在由 4 和 7 组成的数字中找到给定数字的位置
原文:https://www . geesforgeks . org/find-position-给定-number-in-made-4-7/
考虑一系列仅由数字 4 和 7 组成的数字。数列的前几个数字是 4，7，44，47，74，77，444，..等等。给定一个仅由 4，7 位数字构成的数，我们需要找到这个数在这个数列中的位置。 示例:
Input : 7
Output : pos = 2
Input : 444
Output : pos = 7
与以下文章相反: 求数列中只允许有 2 位(4 和 7)的第 n 个元素|集合 2 (log(n)方法)
""
/ \
1(4) 2(7)
/ \ / \
3(44) 4(47) 5(74) 6(77)
/ \ / \ / \ / \
这个想法是基于这样一个事实，即所有偶数位置的数字都有 7 作为最后一个数字，所有奇数位置的数字都有 4 作为最后一个数字。 如果数字是 4，那么它是树的左节点，那么它对应于(pos2)+1。否则右子节点(7)对应于(位置2)+2。
C++
// C++ program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
#include <iostream>
#include <algorithm>
using namespace std;
int findpos(string n)
{
int i = 0, pos = 0;
while (n[i] != '\0') {
// check all digit position
switch (n[i])
{
// if number is left then pos*2+1
case '4':
pos = pos * 2 + 1;
break;
// if number is right then pos*2+2
case '7':
pos = pos * 2 + 2;
break;
}
i++;
}
return pos;
}
// Driver code
int main()
{
// given a number which is constructed
// by 4 and 7 digit only
string n = "774";
cout << findpos(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
import java.util.*;
class GFG {
static int findpos(String n)
{
int k = 0, pos = 0, i = 0;
while (k != n.length()) {
// check all digit position
switch (n.charAt(i)) {
// if number is left then pos*2+1
case '4':
pos = pos * 2 + 1;
break;
// if number is right then pos*2+2
case '7':
pos = pos * 2 + 2;
break;
}
i++;
k++;
}
return pos;
}
// Driver code
public static void main(String[] args)
{
// given a number which is constructed
// by 4 and 7 digit only
String n = "774";
System.out.println(findpos(n));
}
}
// This code is contributed by Sam007.
Python 3
# python program to find position
# of a number in a series of
# numbers with 4 and 7 as the
# only digits.
def findpos(n):
i = 0
j = len(n)
pos = 0
while (i<j):
# check all digit position
# if number is left then
# pos*2+1
if(n[i] == '4'):
pos = pos * 2 + 1
# if number is right then
# pos*2+2
if(n[i] == '7'):
pos = pos * 2 + 2
i= i+1
return pos
# Driver code
# given a number which is constructed
# by 4 and 7 digit only
n = "774"
print(findpos(n))
# This code is contributed by Sam007
C
// C# program to find position of
// a number in a series of numbers
// with 4 and 7 as the only digits.
using System;
class GFG
{
static int findpos(String n)
{
int k = 0, pos = 0, i = 0;
while (k != n.Length) {
// check all digit position
switch (n[i]) {
// if number is left then pos*2+1
case '4':
pos = pos * 2 + 1;
break;
// if number is right then pos*2+2
case '7':
pos = pos * 2 + 2;
break;
}
i++;
k++;
}
return pos;
}
// Driver code
static void Main()
{
// given a number which is constructed
// by 4 and 7 digit only
String n = "774";
Console.Write(findpos(n));
}
}
// This code is contributed by Sam007
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
function findpos($n)
{
$i = 0;
$pos = 0;
while($i < strlen($n)) {
// check all digit position
switch ($n[$i])
{
// if number is left then pos*2+1
case '4':
$pos = $pos * 2 + 1;
break;
// if number is right then pos*2+2
case '7':
$pos = $pos * 2 + 2;
break;
}
$i++;
}
return $pos;
}
// Driver code
// given a number which
// is constructed by 4
// and 7 digit only
$n = "774";
echo findpos($n);
// This code is contributed by Sam007
?>
java 描述语言
<script>
// Javascript program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
function findpos(n)
{
let i = 0;
let pos = 0;
while(i < n.length) {
// check all digit position
switch (n[i])
{
// if number is left then pos*2+1
case '4':
pos = pos * 2 + 1;
break;
// if number is right then pos*2+2
case '7':
pos = pos * 2 + 2;
break;
}
i++;
}
return pos;
}
// Driver code
// given a number which
// is constructed by 4
// and 7 digit only
let n = "774";
document.write(findpos(n));
// This code is contributed by _saurabh_jaiswal
</script>
输出:
13
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