给定整数N ,任务是找到从给定的N个球中选择一些球的方法,以便至少选择一个球。由于该值可能很大,因此请打印模1000000007的值。
例子:
Input: N = 2
Output: 3
The three ways are “*.”, “.*” and “**” where ‘*’ denotes
the chosen ball and ‘.’ denotes the ball which didn’t get chosen.
Input: N = 30000
Output: 165890098
方法:有N个球,每个球都可以选择或不选择。不同配置的总数为2 * 2 * 2 *…* N。我们可以写成2 N。但是必须从答案中减去没有选择球的状态。因此,结果将是(2 N – 1)%1000000007 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
const int MOD = 1000000007;
// Function to return the count of
// ways to choose the balls
int countWays(int n)
{
// Calculate (2^n) % MOD
int ans = 1;
for (int i = 0; i < n; i++) {
ans *= 2;
ans %= MOD;
}
// Subtract the only where
// no ball was chosen
return ((ans - 1 + MOD) % MOD);
}
// Driver code
int main()
{
int n = 3;
cout << countWays(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static int MOD = 1000000007;
// Function to return the count of
// ways to choose the balls
static int countWays(int n)
{
// Calculate (2^n) % MOD
int ans = 1;
for (int i = 0; i < n; i++)
{
ans *= 2;
ans %= MOD;
}
// Subtract the only where
// no ball was chosen
return ((ans - 1 + MOD) % MOD);
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.println(countWays(n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach
MOD = 1000000007
# Function to return the count of
# ways to choose the balls
def countWays(n):
# Return ((2 ^ n)-1) % MOD
return (((2**n) - 1) % MOD)
# Driver code
n = 3
print(countWays(n))
C#
// C# implementation of the approach
using System;
class GFG
{
static int MOD = 1000000007;
// Function to return the count of
// ways to choose the balls
static int countWays(int n)
{
// Calculate (2^n) % MOD
int ans = 1;
for (int i = 0; i < n; i++)
{
ans *= 2;
ans %= MOD;
}
// Subtract the only where
// no ball was chosen
return ((ans - 1 + MOD) % MOD);
}
// Driver code
public static void Main(String[] args)
{
int n = 3;
Console.WriteLine(countWays(n));
}
}
// This code is contributed by Rajput-Ji
输出:
7