赢得比赛所需的最低玩家数

给定N个问题，每个问题有K个选项，其中$1\leq N \leq 1000000000$和$1\leq K \leq 1000000000$ 。任务是确定所有尝试第i个问题的玩家总数$1 \leq i \leq N$无论如何要赢得比赛。您必须最小化播放器总数的总和，并以10 ⁹ +7为模输出。

注意：任何错误答案都会导致玩家被淘汰。

例子：

Input: N = 3, K = 3
Output: 39

Input: N = 5, K = 2
Output: 62

方法：

• To solve Nth question K players are needed.
• To solve (N-1)th question K² players are needed.
• Similarly moving onwards, To solve 1st question K^N players are needed.

因此，我们的问题简化为找到GP项之和K + K ² +…+ K ^N等于$\frac{K(K^{N}-1)}{K-1}$ 。
现在我们可以使用费马小定理以10 ⁹ +7的形式获得所需的答案模。

下面是上述方法的实现：

C++

// C++ program to find minimum players
// required to win the game anyhow
#include 
using namespace std;
#define mod 1000000007
  
// function to calculate (a^b)%(10^9+7).
long long int power(long long int a, long long int b)
{
    long long int res = 1;
    while (b) {
        if (b & 1) {
            res *= a;
            res %= mod;
        }
        b /= 2;
        a *= a;
        a %= mod;
    }
    return res;
}
  
// function to find the minimum required player
long long int minPlayer(long long int n, long long int k)
{
  
    // computing the nenomenator
    long long int num = ((power(k, n) - 1) + mod) % mod;
  
    // computing modulo inverse of denomenator
    long long int den = (power(k - 1, mod - 2) + mod) % mod;
  
    // final result
    long long int ans = (((num * den) % mod) * k) % mod;
  
    return ans;
}
  
// Driver code
int main()
{
  
    long long int n = 3, k = 3;
  
    cout << minPlayer(n, k);
  
    return 0;
}

Java

//Java program to find minimum players
//required to win the game anyhow
public class TYU {
      
    static long  mod =  1000000007;
  
    //function to calculate (a^b)%(10^9+7).
     static long power(long a, long b)
     {
      long res = 1;
      while (b != 0) {
          if ((b & 1) != 0) {
              res *= a;
              res %= mod;
          }
          b /= 2;
          a *= a;
          a %= mod;
      }
      return res;
     }
  
     //function to find the minimum required player
     static long minPlayer(long n, long k)
     {
  
      // computing the nenomenator
      long num = ((power(k, n) - 1) + mod) % mod;
  
      // computing modulo inverse of denomenator
      long den = (power(k - 1, mod - 2) + mod) % mod;
  
      // final result
      long ans = (((num * den) % mod) * k) % mod;
  
      return ans;
     }
  
     //Driver code
    public static void main(String[] args) {
          
         long n = 3, k = 3;
  
         System.out.println(minPlayer(n, k));
    }
}

Python 3

# Python 3 Program  to find minimum players 
# 3 required to win the game anyhow
  
# constant
mod = 1000000007
  
# function to calculate (a^b)%(10^9+7).
def power(a, b) :
    res = 1
  
    while(b) :
        if (b & 1) :
            res *= a
            res %= mod
  
        b //= 2
        a *= a
        a %= mod
  
    return res
  
# function to find the minimum required player 
def minPlayer(n, k) :
  
    # computing the nenomenator 
    num = ((power(k, n) - 1) + mod) % mod
  
    # computing modulo inverse of denomenator 
    den = (power(k - 1,mod - 2) + mod) % mod
  
    # final result 
    ans = (((num * den) % mod ) * k) % mod
  
    return ans
  
# Driver Code
if __name__ == "__main__" :
  
    n, k = 3, 3
  
    print(minPlayer(n, k))
  
# This code is contributed by ANKITRAI1

C#

// C# program to find minimum players
// required to win the game anyhow
using System;
class GFG
{
  
static long mod = 1000000007;
  
// function to calculate (a^b)%(10^9+7).
static long power(long a, long b)
{
long res = 1;
while (b != 0) 
{
    if ((b & 1) != 0) 
    {
        res *= a;
        res %= mod;
    }
    b /= 2;
    a *= a;
    a %= mod;
}
return res;
}
  
// function to find the minimum
// required player
static long minPlayer(long n, long k)
{
  
    // computing the nenomenator
    long num = ((power(k, n) - 1) + mod) % mod;
      
    // computing modulo inverse 
    // of denomenator
    long den = (power(k - 1, mod - 2) + mod) % mod;
      
    // final result
    long ans = (((num * den) % mod) * k) % mod;
      
    return ans;
}
  
// Driver code
public static void Main()
{
    long n = 3, k = 3;
  
    Console.WriteLine(minPlayer(n, k));
}
}
  
// This code is contributed
// by Shashank

输出：

39

时间复杂度： O(log(n))