由取自范围 [L, R] 的 N 个数字形成的不同和的计数

给定三个整数N 、 L和R 。任务是计算使用范围[L, R]中的N个数字形成的不同总和，其中任何数字都可以无限次取。

例子：

Input: N = 2, L = 1, R = 3
Output: 5
Explanation: Generating all distinct combinations of 2 numbers taken from the range [1, 3]
{1, 1} => sum = 2
{1, 2} => sum = 3
{1, 3} => sum = 4
{2, 2} => sum = 4
{2, 3} => sum = 5
{3, 3} => sum = 6
Therefore, there are 5 (2, 3, 4, 5, 6) different sums possible with 2 numbers taken from range [1, 3].

Input: N = 3, L = 1, R = 9
Output: 10

朴素方法：解决给定问题的最简单方法是从范围 [ L , R]和 然后计算这些组合形成的不同总和。

时间复杂度： O((R – L) ^N )
辅助空间： O(1)

有效的方法： 给定的问题可以通过一些观察和使用一些数学来解决。这里可以使用的最小和最大数字分别是L和R。因此，可以形成的最小和最大可能和分别是L*N （所有N个数字都是L）和R*N （所有N个数字都是R），同样，这个范围之间的所有其他和也可以形成。请按照以下步骤解决给定的问题。

• 初始化一个变量，比如minSum = L*N ，以存储可能的最小总和。
• 初始化一个变量maxSum = R*N来存储最大可能的总和。
• 最终答案是[minSum, maxSum]范围内的总数，即(maxSum – minSum + 1) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find total number of
// different sums of N numbers in
// the range [L, R]
int countDistinctSums(int N, int L, int R)
{
 
    // To store minimum possible sum with
    // N numbers with all as L
    int minSum = L * N;
 
    // To store maximum possible sum with
    // N numbers with all as R
    int maxSum = R * N;
 
    // All other numbers in between maxSum
    // and minSum can also be formed so numbers
    // in this range is the final answer
    return maxSum - minSum + 1;
}
 
// Driver Code
int main()
{
    int N = 2, L = 1, R = 3;
    cout << countDistinctSums(N, L, R);
 
    return 0;
}

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG{
 
// Function to find total number of
// different sums of N numbers in
// the range [L, R]
static int countDistinctSums(int N, int L, int R)
{
 
    // To store minimum possible sum with
    // N numbers with all as L
    int minSum = L * N;
 
    // To store maximum possible sum with
    // N numbers with all as R
    int maxSum = R * N;
 
    // All other numbers in between maxSum
    // and minSum can also be formed so numbers
    // in this range is the final answer
    return maxSum - minSum + 1;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2, L = 1, R = 3;
    System.out.print(countDistinctSums(N, L, R));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach
 
# Function to find total number of
# different sums of N numbers in
# the range [L, R]
def countDistinctSums(N, L, R):
 
    # To store minimum possible sum with
    # N numbers with all as L
    minSum = L * N
 
    # To store maximum possible sum with
    # N numbers with all as R
    maxSum = R * N
 
    # All other numbers in between maxSum
    # and minSum can also be formed so numbers
    # in this range is the final answer
    return maxSum - minSum + 1
 
# Driver Code
if __name__ == "__main__":
    N = 2
    L = 1
    R = 3
    print(countDistinctSums(N, L, R))
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find total number of
// different sums of N numbers in
// the range [L, R]
static int countDistinctSums(int N, int L, int R)
{
 
    // To store minimum possible sum with
    // N numbers with all as L
    int minSum = L * N;
 
    // To store maximum possible sum with
    // N numbers with all as R
    int maxSum = R * N;
 
    // All other numbers in between maxSum
    // and minSum can also be formed so numbers
    // in this range is the final answer
    return maxSum - minSum + 1;
}
 
// Driver Code
public static void Main()
{
    int N = 2, L = 1, R = 3;
    Console.Write(countDistinctSums(N, L, R));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find total number of
        // different sums of N numbers in
        // the range [L, R]
        function countDistinctSums(N, L, R) {
 
            // To store minimum possible sum with
            // N numbers with all as L
            let minSum = L * N;
 
            // To store maximum possible sum with
            // N numbers with all as R
            let maxSum = R * N;
 
            // All other numbers in between maxSum
            // and minSum can also be formed so numbers
            // in this range is the final answer
            return maxSum - minSum + 1;
        }
 
        // Driver Code
 
        let N = 2, L = 1, R = 3;
        document.write(countDistinctSums(N, L, R));
 
// This code is contributed by Potta Lokesh
    

输出：

5

时间复杂度： O(1)
辅助空间： O(1)