C语言用递归求斐波那契数，让你发现递归的缺陷和效率瓶颈

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#include <stdio.h>
#include <time.h>
#include <windows.h>
 
// 递归计算斐波那契数
long fibonacci_recursion( int n )
{
    if( n <= 2 )
        return 1;
 
    return fibonacci_recursion(n-1) + fibonacci_recursion(n-2);
}
 
// 迭代计算斐波那契数
long fibonacci_iteration( int n )
{
    long result;
    long previous_result;
    long next_older_result;
 
    result = previous_result = 1;
 
    while( n > 2 ){
        n -= 1;
        next_older_result = previous_result;
        previous_result = result;
        result = previous_result + next_older_result;
    }
    return result;
}
 
int main(){
    int N = 45;
 
    // 递归消耗的时间
    clock_t recursion_start_time = clock();
    long result_recursion = fibonacci_recursion(N);
    clock_t recursion_end_time = clock();
 
    // 迭代消耗的时间
    clock_t iteration_start_time = clock();
    long result_iteration = fibonacci_iteration(N);
    clock_t iteration_end_time = clock();
 
    // 输出递归消耗的时间
    printf("Result of recursion: %ld \nTime: %fseconds",
        fibonacci_recursion(N),
        (double)(recursion_end_time-recursion_start_time)/CLOCKS_PER_SEC
    );
    printf("\n-----------------------\n");
    // 输出迭代消耗的时间
    printf("Result of iteration: %ld \nTime: %fseconds",
        fibonacci_iteration(N),
        (double)(iteration_end_time-iteration_start_time)/CLOCKS_PER_SEC
    );
 
    return 0;
}

#include <stdio.h>
#include <time.h>
#include <windows.h>// 递归计算斐波那契数
long fibonacci_recursion( int n )
{if( n <= 2 )return 1;return fibonacci_recursion(n-1) + fibonacci_recursion(n-2);
}// 迭代计算斐波那契数
long fibonacci_iteration( int n )
{long result;long previous_result;long next_older_result;result = previous_result = 1;while( n > 2 ){n -= 1;next_older_result = previous_result;previous_result = result;result = previous_result + next_older_result;}return result;
}int main(){int N = 45;// 递归消耗的时间clock_t recursion_start_time = clock();long result_recursion = fibonacci_recursion(N);clock_t recursion_end_time = clock();// 迭代消耗的时间clock_t iteration_start_time = clock();long result_iteration = fibonacci_iteration(N);clock_t iteration_end_time = clock();// 输出递归消耗的时间printf("Result of recursion: %ld \nTime: %fseconds",fibonacci_recursion(N),(double)(recursion_end_time-recursion_start_time)/CLOCKS_PER_SEC);printf("\n-----------------------\n");// 输出迭代消耗的时间printf("Result of iteration: %ld \nTime: %fseconds",fibonacci_iteration(N),(double)(iteration_end_time-iteration_start_time)/CLOCKS_PER_SEC);return 0;
}