权重为 W 的 N 位整数个数
给定 N，大于等于 2 的整数的位数和权重 w，任务是找出有 N 位数和权重 w 的整数的个数 注:权重定义为一个整数的连续位数之差。
示例:
Input : N = 2, W = 3
Output : 6
Input : N = 2, W = 4
Output : 5
在上面的例子中，权重等于 3 的 2 位整数的总数可能是 6。就像数字 14 的权重是 3 (4-1)，25、36、47、58、69 的权重是 3。如果我们仔细看，我们会发现这样一个逻辑:如果我们把一个 2 位数的权重增加到 5，那么这些数字的总数可能是 5。对于 2 位数的权重 6，可能的总数将是 4，然后是 3，以此类推。另外，如果我们增加位数。假设 n 等于 3，权重为 3，那么 n 等于 4，权重为 3，那么可能的总数将是 60 和 600，以此类推。 位数|重量— >所有可能的数字
| 2|2 —> 7 | 2|3 —> 6 | 2|4 —> 5 | 2|5 —> 4 | 2|6 —> 3 | 2|7 —> 2 | 2|8 —> 1 | | 3|2 —> 70 | 3|3 —> 60 | 3|4 —> 50 | 3|5 —> 40 | 3|6 —> 30 | 3|7 —> 20 | 3|8 —> 10 | | 4|2 —>700 | 4|3 —>600 | 4|4 —>500 | 4|5 —>400 | 4|6 —>300 | 4|7 —>200 | 4|8 —>100 | 从上表中可以看出，随着位数的增加，权重为“w”的数字数量遵循一种模式，即以 10^(n-2 的倍数变化，其中“n”是位数。 下面是解决这个问题的分步算法: 1. 检查给定的重量是正的还是负的。 2. 如果为正，则从 9 中减去重量。 3. 如果权重为负，则将权重增加到 10，然后更新新的权重。 4. 对于 n 位整数，将 10^(n-2)乘以这个更新的权重。 5. 这将给出满足这个权重的整数个数。 下面是上述方法的实现: ## C++// CPP program to find total possible numbers
// with n digits and weight w
#include <iostream>
#include<cmath>
using namespace std;
// Function to find total possible numbers
// with n digits and weight w
int findNumbers(int n, int w)
{
int x = 0, sum = 0;
// When Weight of an integer is Positive
if (w >= 0 && w <= 8) {
// Subtract the weight from 9
x = 9 - w;
}
// When weight of an integer is negative
else if (w >= -9 && w <= -1) {
// add the weight to 10 to make it positive
x = 10 + w;
}
sum = pow(10, n - 2);
sum = (x * sum);
return sum;
}
// Driver code
int main()
{
int n, w;
// number of digits in an
// integer and w as weight
n = 3, w = 4;
// print the total possible numbers
// with n digits and weight w
cout << findNumbers(n, w);;
return 0;
}
// Java program to find total
// possible numbers with n
// digits and weight w
class GFG
{
// Function to find total
// possible numbers with n
// digits and weight w
static int findNumbers(int n, int w)
{
int x = 0, sum = 0;
// When Weight of an
// integer is Positive
if (w >= 0 && w <= 8)
{
// Subtract the weight from 9
x = 9 - w;
}
// When weight of an
// integer is negative
else if (w >= -9 && w <= -1)
{
// add the weight to 10
// to make it positive
x = 10 + w;
}
sum = (int)Math.pow(10, n - 2);
sum = (x * sum);
return sum;
}
// Driver code
public static void main(String args[])
{
int n, w;
// number of digits in an
// integer and w as weight
n = 3;
w = 4;
// print the total possible numbers
// with n digits and weight w
System.out.println(findNumbers(n, w));
}
}
// This code is contributed
// by ankita_saini
# Python3 program to find total possible
# numbers with n digits and weight w
# Function to find total possible
# numbers with n digits and weight w
def findNumbers(n, w):
x = 0;
sum = 0;
# When Weight of an integer
# is Positive
if (w >= 0 and w <= 8):
# Subtract the weight from 9
x = 9 - w;
# When weight of an integer
# is negative
elif (w >= -9 and w <= -1):
# add the weight to 10 to
# make it positive
x = 10 + w;
sum = pow(10, n - 2);
sum = (x * sum);
return sum;
# Driver code
# number of digits in an
# integer and w as weight
n = 3;
w = 4;
# print the total possible numbers
# with n digits and weight w
print(findNumbers(n, w));
# This code is contributed
# by mits
// C# program to find total possible
// numbers with n digits and weight w
using System;
class GFG
{
// Function to find total possible
// numbers with n digits and weight w
static int findNumbers(int n, int w)
{
int x = 0, sum = 0;
// When Weight of an integer
// is Positive
if (w >= 0 && w <= 8)
{
// Subtract the weight from 9
x = 9 - w;
}
// When weight of an
// integer is negative
else if (w >= -9 && w <= -1)
{
// add the weight to 10
// to make it positive
x = 10 + w;
}
sum = (int)Math.Pow(10, n - 2);
sum = (x * sum);
return sum;
}
// Driver code
static public void Main ()
{
int n, w;
// number of digits in an
// integer and w as weight
n = 3;
w = 4;
// print the total possible numbers
// with n digits and weight w
Console.WriteLine(findNumbers(n, w));
}
}
// This code is contributed by jit_t
<?php
// PHP program to find total possible
// numbers with n digits and weight w
// Function to find total possible
// numbers with n digits and weight w
function findNumbers($n, $w)
{
$x = 0; $sum = 0;
// When Weight of an integer
// is Positive
if ($w >= 0 && $w <= 8)
{
// Subtract the weight from 9
$x = 9 - $w;
}
// When weight of an integer
// is negative
else if ($w >= -9 && $w <= -1)
{
// add the weight to 10 to
// make it positive
$x = 10 + $w;
}
$sum = pow(10, $n - 2);
$sum = ($x * $sum);
return $sum;
}
// Driver code
// number of digits in an
// integer and w as weight
$n = 3; $w = 4;
// print the total possible numbers
// with n digits and weight w
echo findNumbers($n, $w);
// This code is contributed
// by Akanksha Rai
<script>
// Javascript program to find total possible
// numbers with n digits and weight w
// Function to find total possible
// numbers with n digits and weight w
function findNumbers(n, w)
{
let x = 0, sum = 0;
// When Weight of an integer
// is Positive
if (w >= 0 && w <= 8)
{
// Subtract the weight from 9
x = 9 - w;
}
// When weight of an
// integer is negative
else if (w >= -9 && w <= -1)
{
// add the weight to 10
// to make it positive
x = 10 + w;
}
sum = Math.pow(10, n - 2);
sum = (x * sum);
return sum;
}
let n, w;
// number of digits in an
// integer and w as weight
n = 3;
w = 4;
// print the total possible numbers
// with n digits and weight w
document.write(findNumbers(n, w));
</script>
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