javascript - Recursive algorithm problem.

Question

If your boss gives you one yuan on the first day, he will give you double the amount of the previous day every day. That is 1, 2, 4, 8.... . Now, after 30 days, how much money have you received in total?
Requirements: Use recursive functions to implement

Answer 1

The problem is very simple, the key lies in the idea, the following is the code.

/*
·递归
  1. salarySum(n) = salarySum(n-1) + salary(n)
  2. salary(n) = 2 ^ (n-1)

·非递归
  循环
*/

function salary(nthDay){
  return Math.pow(2, nthDay-1)
}

// 递归
function salarySum(nthDay) {
  if (nthDay > 1) {
    return salarySum(nthDay - 1) + salary(nthDay)
  } else {
    return 1
  }
}

// 非递归
function salarySum(nthDay) {
  let day = 1
  let sum = 0
  while (day <= nthDay) {
    sum += salary(day)
    day++
  }
  return sum
}

Answer 2

PHP版本

function recursion($day){
    if($day == 1){
        return 1;
    }else{
        return recursion($day - 1) + pow(2,$day - 1);
    }
}
echo recursion(30);

Answer 3

Haha, shame, I read the wrong question

#不递归的实现方式
def fn(n):
    return 2 ** (n - 1)

#递归的实现方式
def fn1(n):
    return 1 if n <= 1 else fn1(n-1) * 2

Answer 4

#!/usr/bin/env python

def salary(n):
    '''Your salary everyday'''
    if n <= 1:
        return 1
    return 2*salary(n-1)


def money(n):
    '''Total money you get for n days.'''
    if n <= 0:
        return 0
    s = salary(n)  # or s = 2**(n-1)
    m = money(n-1)
    print("day %d: salary[%d] total[%d]" % (n, s, (s+m)))
    return s+m 

money(30)

day 1: salary[1] total[1]
day 2: salary[2] total[3]
day 3: salary[4] total[7]
day 4: salary[8] total[15]
day 5: salary[16] total[31]
day 6: salary[32] total[63]
day 7: salary[64] total[127]
day 8: salary[128] total[255]
day 9: salary[256] total[511]
day 10: salary[512] total[1023]
day 11: salary[1024] total[2047]
day 12: salary[2048] total[4095]
day 13: salary[4096] total[8191]
day 14: salary[8192] total[16383]
day 15: salary[16384] total[32767]
day 16: salary[32768] total[65535]
day 17: salary[65536] total[131071]
day 18: salary[131072] total[262143]
day 19: salary[262144] total[524287]
day 20: salary[524288] total[1048575]
day 21: salary[1048576] total[2097151]
day 22: salary[2097152] total[4194303]
day 23: salary[4194304] total[8388607]
day 24: salary[8388608] total[16777215]
day 25: salary[16777216] total[33554431]
day 26: salary[33554432] total[67108863]
day 27: salary[67108864] total[134217727]
day 28: salary[134217728] total[268435455]
day 29: salary[268435456] total[536870911]
day 30: salary[536870912] total[1073741823]

Answer 5

My idea is: first calculate the current amount, and then pass today's date and amount until the calculation reaches 30 days.

function total(all, day,money){
    var day = day || 1,
        money = money || 0.5;

    money *= 2;
    // console.log( 'day'+day+': '+ money ); // 输出当前的金额
    day++;
    if(day<=all){
        return money + total(all, day, money);
    }
    return money;
}

Then call total():

total(1); // 1
total(2); // 3
total(7); // 127
total(30); // 1073741823