将一个长数字分解为其组成整数时出现奇怪错误

Getting strange errors when breaking up a long number into its component integers

本文关键字：整数 错误 一个 分解 数字      更新时间：2023-09-26

这是针对Project Euler的第8个问题，该问题要求您在这个非常长的数字中找到五个连续数字的最大乘积。

我的代码肯定是NOT最优雅的解决方案，但我确信它应该能工作。它似乎确实适用于小数字，但一旦我测试的数字大于16位，一切都开始分崩离析（例如，子集得到的是e和0以及不同的数字，而不是实际给定数字中的数字。）

function consecProduct(num,sec){
//converts the number to a string
var parser = num.toString();
var numLength = parser.length;
//prepares an array to hold 5 consecutive digits
var pieces = [];
var greatestProduct = 0;
var piecesTogether = 1;
// The outer loop that runs through each set of five digits
for (i=0; i<numLength-4; i++){
    //fills a string with the five digit subset
    var product = parser.substring(sec-5, sec);
    console.log("product "+product);
    //increments subset by 1
    sec++;
    //fills each array position with a each digit from subset
    for(x=0;x<5;x++){
        pieces[x]=product.substring(x,x+1);
        console.log(x + " is "+ pieces[x]);
    }
    //converts each array digit back to an integer
     for(x=0;x<5;x++){
        pieces[x]=parseInt(pieces[x]);
    }
    console.log("hey");
    //gets the product of the subset
    for(x=0;x<5;x++){
        piecesTogether = piecesTogether*pieces[x];
        console.log(pieces[x] + " work " + piecesTogether);
    }
    //updates the greatestProduct
    if ( piecesTogether > greatestProduct ){
        greatestProduct = piecesTogether;
        console.log("great product " + greatestProduct)
    }
    //resets the product for the next subset
    piecesTogether = 1;
}
return greatestProduct;
}
console.log("hey");
consecProduct(111125455578788855,5);

我一直在用Codecademy的scratchpad测试它，也许这是问题的一部分。我上周刚开始学习js，昨天才开始学习这些欧拉问题，所以我可以用各种方法彻底解决这个问题。有什么想法吗？

只要您将数字作为字符串传递，就可以使用您的方法

但大多数浏览器都有一个forEach方法，可以为您简化它。

function eu8(s, n){
    var max= 0, last= 0, A= s.split(''), L= A.length,
    next, temp;
    A.forEach(function(itm, i, A){
        next= i;
        temp= itm;
        while(next<(i+n) && ++next<L) temp*= A[next];
        if(temp> max){
            max= temp;
            last= i;
        }
    });
    return [' Largest product in a sequence of '+n+' digits  totals '+max+
    ','n found at digits #'+last+'-'+(last+n)+' : '+A.slice(last, last+n)];
}
// a shim for old browsers, (not needed with console):
if(!Array.prototype.forEach){
    Array.prototype.forEach= function(fun, scope){
        var T= this, L= T.length, i= 0;
        if(typeof fun== 'function'){
            while(i< L){
                if(i in T){
                    fun.call(scope, T[i], i, T);
                }
                ++i;
            }
        }
        return T;
    }
}
var s= '8383514919085125086820290424163504559356377168995032348562649291222000387486432845620761935475604819050366697920932015432273771435337266340072387705128115575935425014460947570294275818158944549440881025891661096019719598195504110300188717866666358085201663329077618987279717181749021476776048734274617619666392413744636813999541150937273597312043999174331828004915627872035802437409595473241982712379412840772356975718777505301009358387887491501687808639811743258849513533372548739871812190760522789399701735667528924543523146196411626759899045981351660803008793628326225793570101225880141881354855219845587323306406026446646995422604684079629891934580835393600990916331750430169147648113885025045982027652181257767798206409176994378464211282557774833632004180439443121563895765081630408290308927246861936209942841914894036534524282034126702443265629680626122703321065703277654006714223903324966372058553562951193965443957787594408861841150727372912209556865206484636763870595651959623483481581867874';

alert(eu8(s, 5));
//eu8(s, 5)
>>returned value:
 Largest product in a sequence of 5 digits  totals 204120,
 first found at digits #535-540 : 7,5,9,8,9

我最近刚刚开始研究Project Euler，并实现了一个常规的for循环，而不是使用forEach方法。

function largestProduct(n, d){
  var max = 0
  for(var i = 0; i < n.length - d; i++){
    var prod = 1;
    for(var j = 0; j < d; j++){  
      prod *= Number(n.charAt(i+j));
    }
    if(prod > max){
      max = prod;
    }
  }
  return max;
}
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