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Project Ruler 算法练习之除数问题

news 来源：原创 2024/5/8 7:17:35

问题描述：

The fraction ⁴⁹/₉₈ is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that⁴⁹/₉₈ = ⁴/₈, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, ³⁰/₅₀ = ³/₅, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

思路：

1.找到1位数的所有分数

2.找到两位数的分数，把分子分母有重叠的部分去掉，如果相等，存入数组

3.对比两个数组的结果，找到相同的

javascript实现：

(function(){

  var dig = function (m,n){
  var arr = new Array();
  for(j=m+1;j<n;j++){
 for(var i = m;i < j ;i++){
 arr.push({i:i,j:j}); 
}
 } 
return arr;
 }

var filter = function(arr){
var ret  =new Array();
for(var i = 0; i < arr.length ; i++){

var strN = arr[i].i.toString();
var strM = arr[i].j.toString();

if(strN.indexOf(strM[0]) != -1&&strM[0]!='0'){
if(strN[0] == strM[0] && parseInt(strN)/parseInt(strM) == parseInt(strN[1]) / parseInt(strM[1])){ret.push({i:strN[1],j:strM[1],ori:strN+"/"+strM});}
else if(strN[1] == strM[0] && parseInt(strN)/parseInt(strM) == parseInt(strN[0]) / parseInt(strM[1])){ret.push({i:strN[0],j:strM[1],ori:strN+"/"+strM});}
}

else if(strN.indexOf(strM[1]) != -1&&strM[1]!='0'){
if(strN[0] == strM[1]&& parseInt(strN)/parseInt(strM) == parseInt(strN[1]) / parseInt(strM[0])){ret.push({i:strN[1],j:strM[0],ori:strN+"/"+strM});}
else if (strN[0] == strM[1]  && parseInt(strN)/parseInt(strM) == parseInt(strN[1]) / parseInt(strM[0])){ret.push({i:strN[1],j:strM[0],ori:strN+"/"+strM});}
}

}

return ret;
}

var ret1 = dig(1,10);
var ret2= filter(dig(10,100)); 
console.log(ret2);


for(var j = 0 ;j < ret1.length ;j++){
for(var i = 0;i < ret2.length; i++){
//console.log(ret2[i].i + "/" + ret2[i].j + " = " + ret1[j].i + "/" + ret1[j].j);
if(parseInt(ret2[i].i) / parseInt(ret2[i].j) == parseInt(ret1[j].i) / parseInt(ret1[j].j)){
console.log(ret2[i].ori + " = " + ret1[j].i + "/" + ret1[j].j);
}

}   
}


})();