求位计数的幂次加到幂次 B 的和
给定一个整数，数组 A .求 A[i]中每个元素的集合位的和，其幂为 A[i]。 例:
Input: N = 3, A[] = {1, 2, 3}
Output: 10
Explanation:
Set bit of each array element is
1 = 1 set bit,
2 = 1 set bit,
3 = 2 set bit
store each set bit in b[i].
Compute sum of power(b[i], i)
where i is ranging from 1 to n.
that is sum = power(1, 1)+
power(1, 2)+power(2, 3) = 10
Input: N = 4, A[] = {2, 4, 5, 3}
Output: 42
方法: 我们可以使用模幂方法来计算 mod m 下的 a^b，并内置 __builtin_popcount 函数来计算 A[i]的二进制表示中的设置位数。 以下是上述办法的实施情况。
C++
// C++ program for the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate a^b mod m
// using fast-exponentiation method
int fastmod(int base, int exp, int mod)
{
if (exp == 0)
return 1;
else if (exp % 2 == 0) {
int ans = fastmod(base, exp / 2, mod);
return (ans % mod * ans % mod) % mod;
}
else
return (fastmod(base, exp - 1, mod) % mod * base % mod) % mod;
}
// Function to
// calculate sum
int findPowerSum(int n, int ar[])
{
const int mod = 1e9 + 7;
int sum = 0;
// Iterate for all
// values of array A
for (int i = 0; i < n; i++) {
int base = __builtin_popcount(ar[i]);
int exp = ar[i];
// Calling fast-exponentiation and
// appending ans to sum
sum += fastmod(base, exp, mod);
sum %= mod;
}
return sum;
}
// Driver code.
int main()
{
int n = 3;
int ar[] = { 1, 2, 3 };
cout << findPowerSum(n, ar);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG
{
// Function to calculate a^b mod m
// using fast-exponentiation method
static int fastmod(int base, int exp, int mod)
{
if (exp == 0)
return 1;
else if (exp % 2 == 0)
{
int ans = fastmod(base, exp / 2, mod);
return (ans % mod * ans % mod) % mod;
}
else
return (fastmod(base, exp - 1, mod) %
mod * base % mod) % mod;
}
/* Function to get no of set
bits in binary representation
of positive integer n */
static int countSetBits(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Function to calculate sum
static int findPowerSum(int n, int ar[])
{
final int mod = (int)1e9 + 7;
int sum = 0;
// Iterate for all
// values of array A
for (int i = 0; i < n; i++)
{
int base = countSetBits(ar[i]);
int exp = ar[i];
// Calling fast-exponentiation and
// appending ans to sum
sum += fastmod(base, exp, mod);
sum %= mod;
}
return sum;
}
// Driver code.
public static void main (String[] args)
{
int n = 3;
int ar[] = { 1, 2, 3 };
System.out.println(findPowerSum(n, ar));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 program for the above approach
# Function to calculate a^b mod m
# using fast-exponentiation method
def fastmod(base, exp, mod) :
if (exp == 0) :
return 1;
elif (exp % 2 == 0) :
ans = fastmod(base, exp / 2, mod);
return (ans % mod * ans % mod) % mod;
else :
return (fastmod(base, exp - 1, mod)
% mod * base % mod) % mod;
# Function to
# calculate sum
def findPowerSum(n, ar) :
mod = int(1e9) + 7;
sum = 0;
# Iterate for all values of array A
for i in range(n) :
base = bin(ar[i]).count('1');
exp = ar[i];
# Calling fast-exponentiation and
# appending ans to sum
sum += fastmod(base, exp, mod);
sum %= mod;
return sum;
# Driver code.
if __name__ == "__main__" :
n = 3;
ar = [ 1, 2, 3 ];
print(findPowerSum(n, ar));
# This code is contributed by AnkitRai01
C
// C# program for the above approach
using System;
class GFG
{
// Function to calculate a^b mod m
// using fast-exponentiation method
static int fastmod(int baseval, int exp, int mod)
{
if (exp == 0)
return 1;
else if (exp % 2 == 0)
{
int ans = fastmod(baseval, exp / 2, mod);
return (ans % mod * ans % mod) % mod;
}
else
return (fastmod(baseval, exp - 1, mod) %
mod * baseval % mod) % mod;
}
/* Function to get no of set
bits in binary representation
of positive integer n */
static int countSetBits(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Function to calculate sum
static int findPowerSum(int n, int []ar)
{
int mod = (int)1e9 + 7;
int sum = 0;
// Iterate for all
// values of array A
for (int i = 0; i < n; i++)
{
int baseval = countSetBits(ar[i]);
int exp = ar[i];
// Calling fast-exponentiation and
// appending ans to sum
sum += fastmod(baseval, exp, mod);
sum %= mod;
}
return sum;
}
// Driver code.
public static void Main ()
{
int n = 3;
int []ar = { 1, 2, 3 };
Console.WriteLine(findPowerSum(n, ar));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// javascript program for the above approach
// Function to calculate a^b mod m
// using fast-exponentiation method
function fastmod(base , exp , mod)
{
if (exp == 0)
return 1;
else if (exp % 2 == 0)
{
var ans = fastmod(base, exp / 2, mod);
return (ans % mod * ans % mod) % mod;
} else
return (fastmod(base, exp - 1, mod) % mod * base % mod) % mod;
}
/*
* Function to get no of set bits in binary representation of positive integer n
*/
function countSetBits(n)
{
var count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Function to calculate sum
function findPowerSum(n , ar)
{
var mod = parseInt( 1e9 + 7);
var sum = 0;
// Iterate for all
// values of array A
for (i = 0; i < n; i++) {
var base = countSetBits(ar[i]);
var exp = ar[i];
// Calling fast-exponentiation and
// appending ans to sum
sum += fastmod(base, exp, mod);
sum %= mod;
}
return sum;
}
// Driver code
var n = 3;
var ar = [ 1, 2, 3 ];
document.write(findPowerSum(n, ar));
// This code is contributed by gauravrajput1
</script>
Output:
10
时间复杂度: O(nlog(n)) 辅助空间:* O(1)
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