求偶数指数二项式系数之和
给定正整数 n 。任务是找到偶数索引二项式系数的和。即 nC0+nC2+nC4+nC6+nC8+…….. 例:
Input : n = 4
Output : 8
4C0 + 4C2 + 4C4
= 1 + 6 + 1
= 8
Input : n = 6
Output : 32
方法一:(蛮力) 思路是找出所有的二项式系数，只找出偶数索引值的和。
卡片打印处理机（Card Print Processor 的缩写）
// CPP Program to find sum
// of even index term
#include <bits/stdc++.h>
using namespace std;
// Return the sum of
// even index term
int evenSum(int n)
{
int C[n + 1][n + 1];
int i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= min(i, n); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using
// previously stored values
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
// finding sum of even index term.
int sum = 0;
for (int i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
}
// Driver Program
int main()
{
int n = 4;
cout << evenSum(n) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to find sum
// of even index term
import java.io.*;
import java.math.*;
class GFG {
// Return the sum of
// even index term
static int evenSum(int n)
{
int C[][] = new int [n + 1][n + 1];
int i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.min(i, n); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// else Calculate value using
// previously stored values
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
// finding sum of even index term.
int sum = 0;
for (i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
}
// Driver Program
public static void main(String args[])
{
int n = 4;
System.out.println(evenSum(n));
}
}
/*This code is contributed by Nikita Tiwari.*/
计算机编程语言
# Python Program to find sum of even index term
import math
# Return the sum of even index term
def evenSum(n) :
# Creates a list containing n+1 lists,
# each of n+1 items, all set to 0
C = [[0 for x in range(n + 1)] for y in range(n + 1)]
# Calculate value of Binomial Coefficient
# in bottom up manner
for i in range(0, n + 1):
for j in range(0, min(i, n + 1)):
# Base Cases
if j == 0 or j == i:
C[i][j] = 1
# Calculate value using previously
# stored values
else:
C[i][j] = C[i - 1][j - 1] + C[i - 1][j]
# Finding sum of even index term
sum = 0;
for i in range(0, n + 1):
if n % 2 == 0:
sum = sum + C[n][i]
return sum
# Driver method
n = 4
print evenSum(n)
# This code is contributed by 'Gitanjali'.
C
// C# Program to find sum
// of even index term
using System;
class GFG {
// Return the sum of
// even index term
static int evenSum(int n)
{
int [,]C = new int [n + 1,n + 1];
int i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.Min(i, n); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i,j] = 1;
// else Calculate value using
// previously stored values
else
C[i,j] = C[i - 1,j - 1]
+ C[i - 1,j];
}
}
// finding sum of even index term.
int sum = 0;
for (i = 0; i <= n; i += 2)
sum += C[n,i];
return sum;
}
// Driver Program
public static void Main()
{
int n = 4;
Console.WriteLine(evenSum(n));
}
}
/*This code is contributed by vt_m.*/
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to find sum
// of even index term
// Return the sum of
// even index term
function evenSum($n)
{
$C = array(array());
$i; $j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for ($i = 0; $i <= $n; $i++)
{
for ($j = 0; $j <= min($i, $n); $j++)
{
// Base Cases
if ($j == 0 or $j == $i)
$C[$i][$j] = 1;
// Calculate value using
// previously stored values
else
$C[$i][$j] = $C[$i - 1][$j - 1] +
$C[$i - 1][$j];
}
}
// finding sum of even index term.
$sum = 0;
for ( $i = 0; $i <= $n; $i += 2)
$sum += $C[$n][$i];
return $sum;
}
// Driver Code
$n = 4;
echo evenSum($n) ;
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// Javascript Program to find sum
// of even index term
// Return the sum of
// even index term
function evenSum(n)
{
var C = Array.from(Array(n+1),
()=> Array(n+1).fill(0));
var i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= Math.min(i, n); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using
// previously stored values
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
// finding sum of even index term.
var sum = 0;
for (var i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
}
// Driver Program
var n = 4;
document.write( evenSum(n) );
</script>
输出:
8
时间复杂度: O(n 2 )
方法 2:(使用公式)
偶数索引二项式系数之和:
证明:
We know,
(1 + x)n = nC0 + nC1 x + nC2 x2 + ..... + nCn xn
Now put x = -x, we get
(1 - x)n = nC0 - nC1 x + nC2 x2 + ..... + (-1)n nCn xn
Now, adding both the above equation, we get,
(1 + x)n + (1 - x)n = 2 * [nC0 + nC2 x2 + nC4 x4 + .......]
Put x = 1
(1 + 1)n + (1 - 1)n = 2 * [nC0 + nC2 + nC4 + .......]
2n/2 = nC0 + nC2 + nC4 + .......
2n-1 = nC0 + nC2 + nC4 + .......
以下是该方法的实现:
C++
// CPP Program to find sum even indexed Binomial
// Coefficient.
#include <bits/stdc++.h>
using namespace std;
// Returns value of even indexed Binomial Coefficient
// Sum which is 2 raised to power n-1.
int evenbinomialCoeffSum(int n)
{
return (1 << (n - 1));
}
/* Driver program to test above function*/
int main()
{
int n = 4;
printf("%d", evenbinomialCoeffSum(n));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to find sum even indexed
// Binomial Coefficient.
import java.io.*;
class GFG {
// Returns value of even indexed Binomial Coefficient
// Sum which is 2 raised to power n-1.
static int evenbinomialCoeffSum(int n)
{
return (1 << (n - 1));
}
// Driver Code
public static void main(String[] args)
{
int n = 4;
System.out.println(evenbinomialCoeffSum(n));
}
}
// This code is contributed by 'Gitanjali'.
计算机编程语言
# Python program to find sum even indexed
# Binomial Coefficient
import math
# Returns value of even indexed Binomial Coefficient
# Sum which is 2 raised to power n-1.
def evenbinomialCoeffSum( n):
return (1 << (n - 1))
# Driver method
if __name__ == '__main__':
n = 4
print evenbinomialCoeffSum(n)
# This code is contributed by 'Gitanjali'.
C
// C# Program to find sum even indexed
// Binomial Coefficient.
using System;
class GFG
{
// Returns value of even indexed
// Binomial Coefficient Sum which
// is 2 raised to power n-1.
static int evenbinomialCoeffSum(int n)
{
return (1 << (n - 1));
}
// Driver Code
public static void Main()
{
int n = 4;
Console.WriteLine(evenbinomialCoeffSum(n));
}
}
// This code is contributed by 'Vt_m'.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to find sum
// even indexed Binomial
// Coefficient.
// Returns value of even indexed
// Binomial Coefficient Sum which
// is 2 raised to power n-1.
function evenbinomialCoeffSum( $n)
{
return (1 << ($n - 1));
}
// Driver Code
$n = 4;
echo evenbinomialCoeffSum($n);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// JavaScript Program to find sum even indexed
// Binomial Coefficient.
// Returns value of even indexed Binomial Coefficient
// Sum which is 2 raised to power n-1.
function evenbinomialCoeffSum(n)
{
return (1 << (n - 1));
}
// Driver code
let n = 4;
document.write(evenbinomialCoeffSum(n));
// This code is contributed by code_hunt.
</script>
输出:
8
时间复杂度: O(1) 奇指标二项式系数之和 利用上面的结果我们很容易证明奇指标二项式系数之和也是 2 n-1 。
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