码迷,mamicode.com
首页 > 其他好文 > 详细

Project Ruler 算法练习之除数问题

时间:2014-06-20 11:27:23      阅读:158      评论:0      收藏:0      [点我收藏+]

标签:style   class   blog   code   java   com   

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);
}

}   
}


})();


Project Ruler 算法练习之除数问题,布布扣,bubuko.com

Project Ruler 算法练习之除数问题

标签:style   class   blog   code   java   com   

原文地址:http://blog.csdn.net/lan_liang/article/details/28472011

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!