最小重复单位中的 1 的数量
给定一个正整数 N，其单位的数字是 3。找出最小重复单位中可被给定数字 n 整除的 1 的个数。每个单位的位数为 3 的数都有一个重复单位作为它的倍数。repunit 是一个只有一个的数。它的形式是(10n–1)/9。 例:
输入:3 T3】输出: 3 As 3 除以 111，111 是该数的最小重复单位 倍数。所以 111 中的 1 数是 3。 输入: 13 输出: 6
重发单位是 1，11，111，1111，… x 的下一个 repunit 将始终为 x10+1。如果 x repunit 剩下的余数是 r ，那么下一个 repunit 剩下的余数总是 (r10+1)%n.** 由于 repunit 可以很大，所以不需要找 repunit 编号。简单地数一数就会给我们答案。 所以，找出所有 repunit 数的余数，直到余数变为 0。一旦完成，则为使余数为 0 而进行的迭代次数将为 1 的次数。 下面是上述方法的实现:
C++
// CPP program to print the number of 1s in
// smallest repunit multiple of the number.
#include <bits/stdc++.h>
using namespace std;
// Function to find number of 1s in
// smallest repunit multiple of the number
int countOnes(int n)
{
// to store number of 1s in smallest
// repunit multiple of the number.
int count = 1;
// initialize rem with 1
int rem = 1;
// run loop until rem becomes zero
while (rem != 0) {
// rem*10 + 1 here represents
// the repunit modulo n
rem = (rem * 10 + 1) % n;
count++;
}
// when remainder becomes 0
// return count
return count;
}
// Driver Code
int main()
{
int n = 13;
// Calling function
cout << countOnes(n);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print the
// number of 1s in smallest
// repunit multiple of the number.
import java.io.*;
class GFG
{
// Function to find number
// of 1s in smallest repunit
// multiple of the number
static int countOnes(int n)
{
// to store number of 1s
// in smallest repunit
// multiple of the number.
int count = 1;
// initialize rem with 1
int rem = 1;
// run loop until
// rem becomes zero
while (rem != 0)
{
// rem*10 + 1 here
// represents the
// repunit modulo n
rem = (rem * 10 + 1) % n;
count++;
}
// when remainder becomes
// 0 return count
return count;
}
// Driver Code
public static void main (String[] args)
{
int n = 13;
// Calling function
System.out.println(countOnes(n));
}
}
// This code is contributed by akt_mit
Python 3
# Python3 program to print the
# number of 1s in smallest
# repunit multiple of the number.
# Function to find number
# of 1s in smallest repunit
# multiple of the number
def countOnes(n):
# to store number of 1s
# in smallest repunit
# multiple of the number.
count = 1;
# initialize rem with 1
rem = 1;
# run loop until
# rem becomes zero
while (rem != 0):
# rem*10 + 1 here represents
# the repunit modulo n
rem = (rem * 10 + 1) % n;
count = count + 1;
# when remainder becomes
# 0 return count
return count;
# Driver Code
n = 13;
# Calling function
print(countOnes(n));
# This code is contributed by mits
C
// C# program to print the
// number of 1s in smallest
// repunit multiple of the number.
using System;
class GFG
{
// Function to find number
// of 1s in smallest repunit
// multiple of the number
static int countOnes(int n)
{
// to store number of 1s
// in smallest repunit
// multiple of the number.
int count = 1;
// initialize rem with 1
int rem = 1;
// run loop until
// rem becomes zero
while (rem != 0)
{
// rem*10 + 1 here represents
// the repunit modulo n
rem = (rem * 10 + 1) % n;
count++;
}
// when remainder becomes 0
// return count
return count;
}
// Driver Code
static public void Main ()
{
int n = 13;
// Calling function
Console.WriteLine (countOnes(n));
}
}
// This code is contributed by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print the
// number of 1s in smallest
// repunit multiple of the number.
// Function to find number
// of 1s in smallest repunit
// multiple of the number
function countOnes($n)
{
// to store number of 1s
// in smallest repunit
// multiple of the number.
$count = 1;
// initialize rem with 1
$rem = 1;
// run loop until
// rem becomes zero
while ($rem != 0)
{
// rem*10 + 1 here represents
// the repunit modulo n
$rem = ($rem * 10 + 1) % $n;
$count++;
}
// when remainder becomes
// 0 return count
return $count;
}
// Driver Code
$n = 13;
// Calling function
echo countOnes($n);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to print the
// number of 1s in smallest
// repunit multiple of the number.
// Function to find number
// of 1s in smallest repunit
// multiple of the number
function countOnes(n)
{
// to store number of 1s
// in smallest repunit
// multiple of the number.
let count = 1;
// initialize rem with 1
let rem = 1;
// run loop until
// rem becomes zero
while (rem != 0)
{
// rem*10 + 1 here represents
// the repunit modulo n
rem = (rem * 10 + 1) % n;
count++;
}
// when remainder becomes 0
// return count
return count;
}
let n = 13;
// Calling function
document.write(countOnes(n));
// This code is contributed by rameshtravel07.
</script>
Output:
6
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