rsp

(define (e a b c) (if (= c 0) a (e (b a) b (- c 1))))
(define (f a) (lambda (b) (lambda (c)
  (e c b (- ((a (lambda (a) (+ a 1))) 0) 1)))))

rsp

const f = a => b => c => a(d => e => e(d(b)))(f => c)(g => g);

rsp

function f(a) {
  return function (b) {
    return function (c) {
      return a(function (d) {
        return function (e) {
          return e(d(b));
        };
      })(function (f) {
        return c;
      })(function (g) {

rsp

const e = a => b => c => c ? e(b(a))(b)(c - 1) : a;
const f = a => b => c => e(c)(b)(a(a => a + 1)(0) - 1);

rsp

function e(a) {
  return function (b) {
    return function (c) {
      return c ? e(b(a))(b)(c - 1) : a;
    };
  };
}

function f(a) {
  return function (b) {

rsp

const f = x => y => z => {
  let r = z;
  for (let i = 1; i++ < x(a => a + 1)(0); r = y(r));
  return r;
};

rsp

function f(x) {
  return function (y) {
    return function (z) {
      var r = z;
      for (var i = 1; i++ < x(function (a) { return a + 1; })(0); r = y(r));
      return r;
    };
  };
}

rsp

/*

A task for inFullMobile Language Wars: Round 1

TL;DR

We need a function f that takes a and returns b:
b = f(a)

when:

rsp

console.log(count(f(input1))); // should print 0
console.log(count(f(input2))); // should print 1
console.log(count(f(input3))); // should print 9
console.log(count(f(input4))); // should print 99
console.log(count(f(input5))); // should print 999

rsp

console.log(count(input1)); // should print 1
console.log(count(input2)); // should print 2
console.log(count(input3)); // should print 10
console.log(count(input4)); // should print 100
console.log(count(input5)); // should print 1000