斐波那契数列给定范围内的数字总和的最后一位

给定两个表示范围[M，N]的非负整数M，N ，其中M≤N，任务是找到F _M + F _{M + 1} …+ F _N之和的最后一位，其中F _K为斐波那契数列中的^第K^个斐波那契数。

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … 
 

例子：

Input: M = 3, N = 9 
Output: 6 
Explanation: 
We need to find F₃ + F₄ + F₅ + F₆ + F₇ + F₈ + F₉ 
=> 2 + 3 + 5 + 8 + 13 + 21 + 34 = 86. 
Clearly, the last digit of the sum is 6.
Input: M = 3, N = 7 
Output: 1 
Explanation: 
We need to find F₃ + F₄ + F₅ + F₆ + F₇ 
=> 2 + 3 + 5 + 8 + 13 = 31. 
Clearly, the last digit of the sum is 1. 
 

朴素方法：针对此问题的朴素方法是一个一个地找出所有^第K^个斐波那契数的和，其中K处于[M，N]范围内，并最后返回总和的最后一位。此方法的时间复杂度为O(N)，并且该方法对于N的高阶值失败。
高效方法：针对此问题的有效方法是使用皮萨诺时期的概念。

• 这个想法是分别计算(M – 1)和N个斐波那契数之和，然后减去计算值的最后一位。
• 这是因为所有K^个第Fibonacci数之和的最后一位数字，使得K处于[M，N]范围内，等于该范围内所有K^个Fibonacci数之和的最后一位数字之差。 [0，N]以及所有第K^个斐波纳契数的总和，范围为[0，M – 1] 。
• 这些值可以分别在很短的时间内通过皮萨诺时期的概念来计算。
• 让我们了解皮萨诺时期的运作方式。下表说明了前10个斐波纳契数以及在对这些数执行模2运算时获得的值。

i	0	1	2	3	4	5	6	7	8	9	10
Fi	0	1	1	2	3	5	8	13	21	34	55
Fi mod 2	0	1	1	0	1	1	0	1	1	0	10

• 显然，皮萨诺周期(F_为i mod 2)为3，因为011重复本身和长度(011)= 3。
• 现在，让我们观察以下身份：

7 = 2 * 3 + 1 
Dividend = (Quotient × Divisor) + Remainder 
=> F₇ mod 2 = F₁ mod 2 = 1. 
 

• 因此，在给定F _i mod 10的Pisano周期为60的情况下，我们只计算总和直到剩下的余数，而不是计算范围[0，N]中所有数字的总和的最后一位。

下面是上述方法的实现：

C++

// C++ program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
#include
using namespace std;
 
// Calculate the sum of the first
// N Fibonacci numbers using Pisano
// period
long long fib(long long n)
{
     
    // The first two Fibonacci numbers
    long long f0 = 0;
    long long f1 = 1;
 
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
        // Pisano period for % 10 is 60
        long long rem = n % 60;
 
        // Checking the remainder
        if(rem == 0)
           return 0;
 
        // The loop will range from 2 to
        // two terms after the remainder
        for(long long i = 2; i < rem + 3; i++)
        {
           long long f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
         
        long long s = f1 - 1;
        return s;
    }
}
 
// Driver Code
int main()
{
    long long m = 10087887;
    long long n = 2983097899;
 
    long long final = abs(fib(n) - fib(m - 1));
    cout << final % 10 << endl;
}
 
// This code is contributed by Bhupendra_Singh

Java

// Java program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
import java.util.*;
 
class GFG{
 
// Calculate the sum of the first
// N Fibonacci numbers using Pisano
// period
static int fib(long n)
{
     
    // The first two Fibonacci numbers
    int f0 = 0;
    int f1 = 1;
 
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
         
        // Pisano period for % 10 is 60
        int rem = (int) (n % 60);
 
        // Checking the remainder
        if(rem == 0)
        return 0;
 
        // The loop will range from 2 to
        // two terms after the remainder
        for(int i = 2; i < rem + 3; i++)
        {
           int f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
         
        int s = f1 - 1;
        return s;
    }
}
 
// Driver Code
public static void main(String args[])
{
    int m = 10087887;
    long n = 2983097899L;
    int Final = (int)Math.abs(fib(n) -
                              fib(m - 1));
     
    System.out.println(Final % 10);
}
}
 
// This code is contributed by AbhiThakur

Python3

# Python3 program to calculate
# Last Digit of the sum of the
# Fibonacci numbers from M to N
 
# Calculate the sum of the first
# N Fibonacci numbers using Pisano
# period
def fib(n):
 
    # The first two Fibonacci numbers
    f0 = 0
    f1 = 1
 
    # Base case
    if (n == 0):
        return 0
    if (n == 1):
        return 1
    else:
 
        # Pisano Period for % 10 is 60
        rem = n % 60
 
        # Checking the remainder
        if(rem == 0):
            return 0
 
        # The loop will range from 2 to
        # two terms after the remainder
        for i in range(2, rem + 3):
            f =(f0 + f1)% 60
            f0 = f1
            f1 = f
 
        s = f1-1
        return(s)
 
# Driver code
if __name__ == '__main__':
     
    m = 10087887
    n = 2983097899
 
    final = fib(n)-fib(m-1)
 
    print(final % 10)

C#

// C# program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
using System;
 
class GFG{
 
// Calculate the sum of the first
// N fibonacci numbers using Pisano
// period
static int fib(long n)
{
     
    // The first two fibonacci numbers
    int f0 = 0;
    int f1 = 1;
 
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
         
        // Pisano period for % 10 is 60
        int rem = (int)(n % 60);
 
        // Checking the remainder
        if(rem == 0)
           return 0;
 
        // The loop will range from 2 to
        // two terms after the remainder
        for(int i = 2; i < rem + 3; i++)
        {
           int f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
         
        int s = f1 - 1;
        return s;
    }
}
 
// Driver Code
public static void Main()
{
    int m = 10087887;
    long n = 2983097899L;
    int Final = (int)Math.Abs(fib(n) -
                              fib(m - 1));
     
    Console.WriteLine(Final % 10);
}
}
 
// This code is contributed by Code_Mech

Javascript

// javascript program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
 
    // Calculate the sum of the first
    // N Fibonacci numbers using Pisano
    // period
    function fib(n) {
 
        // The first two Fibonacci numbers
        var f0 = 0;
        var f1 = 1;
 
        // Base case
        if (n == 0)
            return 0;
        if (n == 1)
            return 1;
        else {
 
            // Pisano period for % 10 is 60
            var rem = parseInt( (n % 60));
 
            // Checking the remainder
            if (rem == 0)
                return 0;
 
            // The loop will range from 2 to
            // two terms after the remainder
            for (i = 2; i < rem + 3; i++) {
                var f = (f0 + f1) % 60;
                f0 = f1;
                f1 = f;
            }
 
            var s = f1 - 1;
            return s;
        }
    }
 
    // Driver Code
    var m = 10087887;
    var n = 2983097899;
    var Final = parseInt( Math.abs(fib(n) - fib(m - 1)));
 
    document.write(Final % 10);
 
// This code is contributed by Rajput-Ji

输出：

5

时间复杂度： O(1) ，因为此代码对于任何输入数字都将运行近60次。