青蛙上台阶

//第一次尝试：
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>

//f(1) = 1;
//f(2) = f(1) + 1 = f(1) + f(1) = 2 * f(1) = 2;
//f(3) = f(2) + f(1) + 1 = f(2) + f(1) + f(1)= f(2) + 2 * f(1) = f(2) + f(2) = 2 * f(2) = 2 * 2;
//f(k) = f(k - 1) + f(k - 2) + … + f(2) + f(1) + 1
//  = f(k - 1) + f(k - 2) + … + f(3) + f(2) + f(1) + f(1)
//  = f(k - 1) + f(k - 2) + … + f(3) + f(2) + f(2）
//    = f(k - 1) + f(k - 2) + f(k - 3) + … + f(3) + f(3)
//    = 2 * f(k - 1)

int FrogJumps(int n) {
    if (n == 1) {
        return 1;
    }
    else {
        return 2*FrogJumps(n - 1);
    }
}

int main() {
    int n = 0;
    printf("输入青蛙一次最多可以跳几个台阶：\n");
    scanf("%d", &n);
    int m = n;
    int ret = FrogJumps(n,m);
    printf("%d\n", ret);
    return 0;
}
//本题看起来很可怕，但是仔细算一算，还是可以找到递推公式的

//f(k) = 2 * f(k - 1)   条件：k>2