求给定数字开始的 k 位数和结束的 l 位数的平均值
原文:https://www . geesforgeks . org/find-从给定数字的开头和结尾找到 k 位数的平均值/
给定三个整数 N 、 K 和 L 。任务是找到给定编号 N 的前 K 位和后 L 位的平均值，没有任何数字重叠。 示例:
输入: N = 123456，K = 2，L = 3 输出: 3.0 前 K 位之和为 1 + 2 = 3 后 L 位之和为 4 + 5 + 6 = 15 平均值= (3 + 15) / (2 + 3) = 18 / 5 = 3 输入: N = 456966，K = 1，L = 1 【T11
方法:如果 n 中的位数小于 (K + L) ，则不可能找到没有位数重叠的平均值，打印 -1 。如果不是这样，求 N 的最后 L 位的和，存入一个变量，比如 sum1 ，然后求 N 的第一个 K 位的和，存入 sum2 。现在，将平均值打印为 (sum1 + sum2) / (K + L) 。 以下是上述办法的实施:
C++
// implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count
// of digits in num
int countDigits(int num)
{
int cnt = 0;
while (num > 0)
{
cnt++;
num /= 10;
}
return cnt;
}
// Function to return the sum
// of first n digits of num
int sumFromStart(int num, int n, int rem)
{
// Remove the unnecessary digits
num /= ((int)pow(10, rem));
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to return the sum
// of the last n digits of num
int sumFromEnd(int num, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
float getAverage(int n, int k, int l)
{
// If the average can't be calculated without
// using the same digit more than once
int totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
int sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
int sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ((float)(sum1 + sum2) /
(float)(k + l));
}
// Driver code
int main()
{
int n = 123456, k = 2, l = 3;
cout << getAverage(n, k, l);
return 0;
}
// This code is contributed by PrinciRaj1992
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the count
// of digits in num
public static int countDigits(int num)
{
int cnt = 0;
while (num > 0) {
cnt++;
num /= 10;
}
return cnt;
}
// Function to return the sum
// of first n digits of num
public static int sumFromStart(int num, int n, int rem)
{
// Remove the unnecessary digits
num /= ((int)Math.pow(10, rem));
int sum = 0;
while (num > 0) {
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to return the sum
// of the last n digits of num
public static int sumFromEnd(int num, int n)
{
int sum = 0;
for (int i = 0; i < n; i++) {
sum += (num % 10);
num /= 10;
}
return sum;
}
public static float getAverage(int n, int k, int l)
{
// If the average can't be calculated without
// using the same digit more than once
int totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
int sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
int sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ((float)(sum1 + sum2) / (float)(k + l));
}
// Driver code
public static void main(String args[])
{
int n = 123456, k = 2, l = 3;
System.out.print(getAverage(n, k, l));
}
}
Python 3
# implementation of the approach
from math import pow
# Function to return the count
# of digits in num
def countDigits(num):
cnt = 0
while (num > 0):
cnt += 1
num //= 10
return cnt
# Function to return the sum
# of first n digits of num
def sumFromStart(num, n, rem):
# Remove the unnecessary digits
num //= pow(10, rem)
sum = 0
while (num > 0):
sum += (num % 10)
num //= 10
return sum
# Function to return the sum
# of the last n digits of num
def sumFromEnd(num, n):
sum = 0
for i in range(n):
sum += (num % 10)
num //= 10
return sum
def getAverage(n, k, l):
# If the average can't be calculated without
# using the same digit more than once
totalDigits = countDigits(n)
if (totalDigits < (k + l)):
return -1
# Sum of the last l digits of n
sum1 = sumFromEnd(n, l)
# Sum of the first k digits of n
# (totalDigits - k) must be removed from the
# end of the number to get the remaining
# k digits from the beginning
sum2 = sumFromStart(n, k, totalDigits - k)
# Return the average
return (sum1 + sum2) / (k + l)
# Driver code
if __name__ == '__main__':
n = 123456
k = 2
l = 3
print(getAverage(n, k, l))
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of digits in num
public static int countDigits(int num)
{
int cnt = 0;
while (num > 0)
{
cnt++;
num /= 10;
}
return cnt;
}
// Function to return the sum
// of first n digits of num
public static int sumFromStart(int num,
int n, int rem)
{
// Remove the unnecessary digits
num /= ((int)Math.Pow(10, rem));
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to return the sum
// of the last n digits of num
public static int sumFromEnd(int num, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
public static float getAverage(int n, int k, int l)
{
// If the average can't be calculated without
// using the same digit more than once
int totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
int sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
int sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ((float)(sum1 + sum2) /
(float)(k + l));
}
// Driver code
public static void Main(String []args)
{
int n = 123456, k = 2, l = 3;
Console.WriteLine(getAverage(n, k, l));
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// javascript implementation of the approach
// Function to return the count
// of digits in num
function countDigits(num)
{
var cnt = 0;
while (num > 0)
{
cnt++;
num = parseInt(num/10);
}
return cnt;
}
// Function to return the sum
// of first n digits of num
function sumFromStart(num, n, rem)
{
// Remove the unnecessary digits
num = (parseInt( num/Math.pow(10, rem)));
var sum = 0;
while (num > 0)
{
sum += (num % 10);
num = parseInt(num/10);
}
return sum;
}
// Function to return the sum
// of the last n digits of num
function sumFromEnd(num , n)
{
var sum = 0;
for (i = 0; i < n; i++)
{
sum += (num % 10);
num = parseInt(num/10);
}
return sum;
}
function getAverage(n , k , l) {
// If the average can't be calculated without
// using the same digit more than once
var totalDigits = countDigits(n);
if (totalDigits < (k + l))
return -1;
// Sum of the last l digits of n
var sum1 = sumFromEnd(n, l);
// Sum of the first k digits of n
// (totalDigits - k) must be removed from the
// end of the number to get the remaining
// k digits from the beginning
var sum2 = sumFromStart(n, k, totalDigits - k);
// Return the average
return ( (sum1 + sum2) / (k + l));
}
// Driver code
var n = 123456, k = 2, l = 3;
document.write(getAverage(n, k, l));
// This code is contributed by Rajput-Ji
</script>
Output:
3.6
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