n 在由 2、3、5 组成的数字中的位置& 7
原文:https://www . geesforgeks . org/position-n-in-numbers-made-2-3-5-7/
考虑一系列仅由数字 2、3、5、7(素数)组成的数字。数列中的前几个数字是 2、3、5、7、22、23、25、27、32、33、35、37、52、53、55、57..等等。给定一个仅由 2，3，5，7 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 : 22
Output : 5
22 is 5th number in series 2, 3, 5, 7, 22, ...
Input : 777
Output : 84
与以下文章相反: 只求质数(2、3、5、7)组成的第 n 个数
""
/ | | \
1(2) 2(3) 3(5) 4(7)
/ | | \ / | | \ / | | \ / | | \
5(22) 6(23) 7(25) 8(27) 9(32)10(33)11(35)12(37) 13(52) 14(53) 15(55) 16(57) 17(72) 18(73) 19(75) 20(77)
/||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\ /||\
如果数字为 2，则位于位置2+1 上；如果数字为 3，则位于位置2+2 上；如果数字为 5，则位于位置2+3 上；如果数字为 7，则位于位置2+4 上；这里，位置是大于或等于 0 的整数。
C++
#include <algorithm>
#include <iostream>
using namespace std;
int findpos(string n)
{
int pos = 0;
for (int i = 0; n[i] != '\0'; i++) {
switch (n[i]) {
// If number is 2 then it is
// on the position pos*2+1
case '2':
pos = pos * 4 + 1;
break;
// If number is 3 then it is
// on the position pos*2+2
case '3':
pos = pos * 4 + 2;
break;
// If number is 5 then it is
// on the position pos*2+3
case '5':
pos = pos * 4 + 3;
break;
// If number is 7 then it is
// on the position pos*2+4
case '7':
pos = pos * 4 + 4;
break;
}
}
return pos;
}
// Driver code
int main()
{
string n = "777";
cout << findpos(n);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program position of n among
// the numbers made of 2, 3, 5 & 7
class GFG
{
static int findpos(String n)
{
int pos = 0;
for (int i = 0; i < n.length(); i++)
{
switch (n.charAt(i))
{
// If number is 2 then it is
// on the position pos*2+1
case '2':
pos = pos * 4 + 1;
break;
// If number is 3 then it is
// on the position pos*2+2
case '3':
pos = pos * 4 + 2;
break;
// If number is 5 then it is
// on the position pos*2+3
case '5':
pos = pos * 4 + 3;
break;
// If number is 7 then it is
// on the position pos*2+4
case '7':
pos = pos * 4 + 4;
break;
}
}
return pos;
}
// Driver code
public static void main(String args[])
{
String n = "777";
System.out.println( findpos(n));
}
}
// This code is contributed
// by Arnab Kundu
Python 3
def findpos(n):
pos = 0
for i in n:
# If number is 2 then it is
# on the position pos*2+1
if i == '2':
pos = pos * 4 + 1
# If number is 3 then it is
# on the position pos*2+2
elif i == '3':
pos = pos * 4 + 2
# If number is 5 then it is
# on the position pos*2+3
elif i == '5':
pos = pos * 4 + 3
# If number is 7 then it is
# on the position pos*2+4
elif i == '7':
pos = pos * 4 + 4
return pos
# Driver code
n = "777"
print (findpos(n))
# This code is contributed by vishal.
C
// C# Program position of n among
// the numbers made of 2, 3, 5 & 7
using System;
class GFG
{
static int findpos(String n)
{
int pos = 0;
for (int i = 0; i < n.Length; i++)
{
switch (n[i])
{
// If number is 2 then it is
// on the position pos*2+1
case '2':
pos = pos * 4 + 1;
break;
// If number is 3 then it is
// on the position pos*2+2
case '3':
pos = pos * 4 + 2;
break;
// If number is 5 then it is
// on the position pos*2+3
case '5':
pos = pos * 4 + 3;
break;
// If number is 7 then it is
// on the position pos*2+4
case '7':
pos = pos * 4 + 4;
break;
}
}
return pos;
}
// Driver code
public static void Main(String[] args)
{
String n = "777";
Console.WriteLine( findpos(n));
}
}
// This code contributed by Rajput-Ji
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program position of n among
// the numbers made of 2, 3, 5 & 7
function findpos($n)
{
$pos = 0;
for ($i = 0; isset($n[$i]) != NULL; $i++)
{
switch ($n[$i])
{
// If number is 2 then it is
// on the position pos*2+1
case '2':
$pos = $pos * 4 + 1;
break;
// If number is 3 then it is
// on the position pos*2+2
case '3':
$pos = $pos * 4 + 2;
break;
// If number is 5 then it is
// on the position pos*2+3
case '5':
$pos = $pos * 4 + 3;
break;
// If number is 7 then it is
// on the position pos*2+4
case '7':
$pos = $pos * 4 + 4;
break;
}
}
return $pos;
}
// Driver Code
$n = "777";
echo findpos($n);
// This code is contributed by nitin mittal.
?>
java 描述语言
<script>
// Javascript Program position of n among
// the numbers made of 2, 3, 5 & 7
function findpos(n)
{
var pos = 0;
for (i = 0; i < n.length; i++)
{
switch (n.charAt(i))
{
// If number is 2 then it is
// on the position pos*2+1
case '2':
pos = pos * 4 + 1;
break;
// If number is 3 then it is
// on the position pos*2+2
case '3':
pos = pos * 4 + 2;
break;
// If number is 5 then it is
// on the position pos*2+3
case '5':
pos = pos * 4 + 3;
break;
// If number is 7 then it is
// on the position pos*2+4
case '7':
pos = pos * 4 + 4;
break;
}
}
return pos;
}
// Driver code
var n = "777";
document.write( findpos(n));
// This code is contributed by Amit Katiyar
</script>
Output:
84
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