數學 × 程式編寫比賽 (第八回) -- 我的解法

[vincentyip]

這次我又參加了 @kenchung 的「數學 × 程式編寫比賽 (第八回)」了，真的非常益智和有趣！

帖文可到這裡看一看：https://steemit.com/contest/@kenchung/question-mathematics-programming-competition-8

要解決這條問題，我們首先要思考有那些數字在旋轉後會成為一個有意義的數字。我們可以由0-9999之中逐個數字檢查，嘗試將那數字旋轉，如果旋轉後是另一個有意義而且比自己大的數字，便在那個數字作標記。

所以檢查完成後，沒有標記的數字的數量便是我們的答案了！

程式如下：

var alts=[0,1,5,-1,-1,2,9,-1,8,6];

var log=[];
var x=0;

for(var i=0; i<10000; i++)
    log[i]=false;

for(var i=0; i<10000; i++){

    if(log[i]) continue;
    log[i]=true;
    x++;

    var b=i.toString();

    if(i<1000) b='0'+b;
    if(i<100) b='0'+b;
    if(i<10) b='0'+b;

    var q='';

    for(j=0; j<4; j++)
        q=alts[b[j]]+q;

    if(q.indexOf(-1)===-1)
        log[parseInt(q)]=true;

    console.log(b,q);
}
console.log(x);

希望可以得獎啦 :D

Comments

[kenchung]

程式看來不錯～

[yasirgawad]

good post dear friend
thank you for sharing

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[duongtuyethai]

不错哦
竖大拇指