求一js判断能否被除尽方法

[longge424]

输入两个数 判断是否能被除尽 如 1/3不能被除尽 1/4可以被除尽 用js如何验证

[inetfuture]

js算？js本身就有精度的bug 怎么算的了……关注 帮顶

[王集鹄]

之前玩过这个算法题：算法题，分数变小数，路过有分

function fraction2decimal(m, n) {
	var num = {};
	var i = 0;
	var x = parseInt(m / n);
	m = m % n;
	var result = "";
	while (m != 0 && !(m in num)) {
		num[m] = i++;
		result += parseInt(m * 10 / n);
		m = m * 10 % n;  
	}
	if (m) {
		var index = num[m];
		return x + ". " + result.substring(0, index) + "( " + result.substring(index) + ") ";
	}
	return x + ". " + result;
}

document.open();
var data = ["1/3", "1/4", "1/7", "1/9", "1/21", "3/13", "17/11", "8/29"];
for (var i = 0; i < data.length; i++) {
	var m = data[i].match(/(\d+)\/(\d+)/);
	if (!m) continue;
	document.write(data[i]);
	document.write("=");
	document.write(fraction2decimal(+m[1], +m[2]));	
	document.write("<br/>");
}
document.close();

输出

1/3=0.(3) 
1/4=0. 25
1/7=0.(142857) 
1/9=0.(1) 
1/21=0.(047619) 
3/13=0.(230769) 
17/11=1.(54) 
8/29=0.(2758620689655172413793103448) 

楼主的需求只需要判断最后m是不是零。
零就是可以除尽。

return !m;

[q107770540]

清洁工 V5

[-晴天]

判断取余不为零就行了.