[Answer] Mathematics × Programming Competition #8 [答案] 數學 × 程式編寫比賽 ...

@kenchung 's Contest is fun, as always, thank you!

This puzzle is simple to understand.

If you flip this over 180 degree, you will get

So how many numbers do you need to represent from 0000 to 9999?

First, we need a function to tell if a digit can be flipped... if not, e.g. like 7, we need to return false.

function getRevDig(x) {
    switch (x) {
        case 0: return 0;
        case 1: return 1;
        case 2: return 2;
        case 5: return 5;
        case 6: return 9;
        case 8: return 8;
        case 9: return 6;
        default: return false;
    }
}

Now, given a four digit number, we need to return its flipped version, to get each digit, you can use module operator and division.

function getRev(x) {
    var a = x % 10;
    a = getRevDig(a);
    if (a === false) return false;
    var b = Math.floor(x / 10) % 10;
    b = getRevDig(b);
    if (b === false) return false;
    var c = Math.floor(x / 100) % 10;
    c = getRevDig(c);
    if (c === false) return false;  
    var d = Math.floor(x / 1000);
    d = getRevDig(d);
    if (d === false) return false;  
    return (a * 1000 + b * 100 + c * 10 + d);
}

The rest is straightforward, just to check and mark the number and its rotated version in a dictionary-like data structure (or array).


var total = 0;
var a = {};
for (var i = 0; i <= 9999; ++i) {
    if (!(i in a)) {
        total ++;
        a[i] = 1;
        var x = getRev(i);
        if (x !== false) {
            a[x] = 1;
        }
    }
}
console.log(total);

That should give you the correct answer.

Designed by @nicolemoker

@kenchung 的数学及编程比赛真好玩，可以练练脑，练练手，防止老年痴呆。

这次比赛不难，就是把一个4个数字的电子屏倒过来，

就成这样：

问题是我们需要多少块（每块倒着可以用）

首先，写一个函数，用于返回一个数字是否可以倒过来，像7这种倒过来没有意义的就可以返回FALSE区分。

function getRevDig(x) {
    switch (x) {
        case 0: return 0;
        case 1: return 1;
        case 2: return 2;
        case 5: return 5;
        case 6: return 9;
        case 8: return 8;
        case 9: return 6;
        default: return false;
    }
}

给定一个4位数，计算倒过来的数字，如果倒过来无意义，返回FALSE。

function getRev(x) {
    var a = x % 10;
    a = getRevDig(a);
    if (a === false) return false;
    var b = Math.floor(x / 10) % 10;
    b = getRevDig(b);
    if (b === false) return false;
    var c = Math.floor(x / 100) % 10;
    c = getRevDig(c);
    if (c === false) return false;  
    var d = Math.floor(x / 1000);
    d = getRevDig(d);
    if (d === false) return false;  
    return (a * 1000 + b * 100 + c * 10 + d);
}

然后就是从0到9999，每次检查是否已经出现了，否则统计加1，然后把倒过来的数字也标记上了。


var total = 0;
var a = {};
for (var i = 0; i <= 9999; ++i) {
    if (!(i in a)) {
        total ++;
        a[i] = 1;
        var x = getRev(i);
        if (x !== false) {
            a[x] = 1;
        }
    }
}
console.log(total);

这样就能打印出正确的答案了，希望能获奖！

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[Answer] Mathematics × Programming Competition #8 [答案] 數學 × 程式編寫比賽 (第八回)