给定一个数字n。求所有n到n的数字之和，其中2位已设置。
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例如：
Input : 10Output : 333 + 5 + 6 + 9 + 10 = 33Input : 100Output : 762
天真的方法： 找到每一个设置了2位的数字，最大为n。如果设置了2位，则将其添加到总和中。
C++
// CPP program to find sum of numbers // upto n whose 2 bits are set #include <bits/stdc++.h> using namespace std; // To count number of set bits int countSetBits( int n) { int count = 0; while (n) { n &= (n - 1); count++; } return count; } // To calculate sum of numbers int findSum( int n) { int sum = 0; // To count sum of number // whose 2 bit are set for ( int i = 1; i <= n; i++) if (countSetBits(i) == 2) sum += i; return sum; } // Driver program to test above function int main() { int n = 10; cout << findSum(n); return 0; } |
JAVA
// Java program to find sum of numbers // upto n whose 2 bits are set public class Main { // To count number of set bits static int countSetBits( int n) { int count = 0 ; while (n > 0 ) { n &= (n - 1 ); count++; } return count; } // To calculate sum of numbers static int findSum( int n) { int sum = 0 ; // To count sum of number // whose 2 bit are set for ( int i = 1 ; i <= n; i++) if (countSetBits(i) == 2 ) sum += i; return sum; } // Driver program to test above function public static void main(String[] args) { int n = 10 ; System.out.println(findSum(n)); } } |
Python3
# Python program to find # sum of numbers # upto n whose 2 bits are set # To count number of set bits def countSetBits(n): count = 0 while (n): n = n & (n - 1 ) count = count + 1 return count # To calculate sum of numbers def findSum(n): sum = 0 # To count sum of number # whose 2 bit are set for i in range ( 1 ,n + 1 ): if (countSetBits(i) = = 2 ): sum = sum + i return sum # Driver code n = 10 print (findSum(n)) # This code is contributed # by Anant Agarwal. |
C#
// C# program to find sum of // numbers upto n whose 2 // bits are set using System; class GFG { // To count number // of set bits static int countSetBits( int n) { int count = 0; while (n > 0) { n = n & (n - 1); count++; } return count; } // To calculate // sum of numbers static int findSum( int n) { int sum = 0; // To count sum of number // whose 2 bit are set for ( int i = 1; i <= n; i++) if (countSetBits(i) == 2) sum += i; return sum; } // Driver Code static public void Main () { int n = 10; Console.WriteLine(findSum(n)); } } // This code is contributed by aj_36 |
PHP
<?php // PHP program to find sum of numbers // upto n whose 2 bits are set // To count number of set bits function countSetBits( $n ) { $count = 0; while ( $n ) { $n &= ( $n - 1); $count ++; } return $count ; } // To calculate sum of numbers function findSum( $n ) { $sum = 0; // To count sum of number // whose 2 bit are set for ( $i = 1; $i <= $n ; $i ++) if (countSetBits( $i ) == 2) $sum += $i ; return $sum ; } // Driver Code $n = 10; echo findSum( $n ); // This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript program to find sum of numbers // upto n whose 2 bits are set // To count number of set bits function countSetBits(n) { let count = 0; while (n) { n &= (n - 1); count++; } return count; } // To calculate sum of numbers function findSum(n) { let sum = 0; // To count sum of number // whose 2 bit are set for (let i = 1; i <= n; i++) if (countSetBits(i) == 2) sum += i; return sum; } // Driver program to test above function let n = 10; document.write(findSum(n)); // This code is contributed by Mayank Tyagi </script> |
输出：
33
有效方法： 设置了2位的数字的形式是2^x+2^y，这个数字小于n。因此，我们必须只找到n范围内的数字，其形式是2^i+2^j，其中i>0，2^i
C++
// C++ program to find sum of numbers // upto n whose 2 bits are set #include <bits/stdc++.h> using namespace std; // To calculate sum of numbers int findSum( int n) { int sum = 0; // Find numbers whose 2 bits are set for ( int i = 1; (1 << i) < n; i++) { for ( int j = 0; j < i; j++) { int num = (1 << i) + (1 << j); // If number is greater then n // we don't include this in sum if (num <= n) sum += num; } } // Return sum of numbers return sum; } // Driver program to test findSum() int main() { int n = 10; cout << findSum(n); return 0; } |
JAVA
// Java program to find sum of numbers // upto n whose 2 bits are set public class Main { // To calculate sum of numbers static int findSum( int n) { int sum = 0 ; // Find numbers whose 2 bits are set for ( int i = 1 ; 1 << i < n; i++) { for ( int j = 0 ; j < i; j++) { int num = ( 1 << i) + ( 1 << j); // If number is greater then n // we don't include this in sum if (num <= n) sum += num; } } // Return sum of numbers return sum; } // Driver program to test findSum() public static void main(String[] args) { int n = 10 ; System.out.println(findSum(n)); } } |
Python3
# Python3 program to find sum of # numbers upto n whose 2 bits are set # To calculate sum of numbers def findSum(n) : sum = 0 # Find numbers whose 2 # bits are set i = 1 while (( 1 << i) < n ) : for j in range ( 0 , i) : num = ( 1 << i) + ( 1 << j) # If number is greater then n # we don't include this in sum if (num < = n) : sum + = num i + = 1 # Return sum of numbers return sum # Driver Code n = 10 print (findSum(n)) # This code is contributed # by Smitha |
C#
// C# program to find sum of numbers // upto n whose 2 bits are set using System; public class main { // To calculate sum of numbers static int findSum( int n) { int sum = 0; // Find numbers whose 2 bits are set for ( int i = 1; 1 << i < n; i++) { for ( int j = 0; j < i; j++) { int num = (1 << i) + (1 << j); // If number is greater then n // we don't include this in sum if (num <= n) sum += num; } } // Return sum of numbers return sum; } // Driver Code public static void Main(String []args) { int n = 10; Console.WriteLine(findSum(n)); } } // This Code is contributed by vt_m. |
PHP
<?php <?php // PHP program to find sum of numbers // upto n whose 2 bits are set // To calculate sum of numbers function findSum( $n ) { $sum = 0; // Find numbers whose 2 bits are set for ( $i = 1; (1 << $i ) < $n ; $i ++) { for ( $j = 0; $j < $i ; $j ++) { $num = (1 << $i ) + (1 << $j ); // If number is greater then n // we don't include this in sum if ( $num <= $n ) $sum += $num ; } } // Return sum of numbers return $sum ; } // Driver Code $n = 10; echo findSum( $n ); // This code is contributed by Ajit ?> |
Javascript
<script> // Javascript program to find sum of numbers // upto n whose 2 bits are set // To calculate sum of numbers function findSum(n) { let sum = 0; // Find numbers whose 2 bits are set for (let i = 1; 1 << i < n; i++) { for (let j = 0; j < i; j++) { let num = (1 << i) + (1 << j); // If number is greater then n // we don't include this in sum if (num <= n) sum += num; } } // Return sum of numbers return sum; } let n = 10; document.write(findSum(n)); </script> |
输出：
33
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