啤酒2元一瓶,四个瓶盖可换一瓶啤酒,2个空瓶也可换一瓶啤酒,10元最多可以喝几瓶，要用js实现
巴扎黑2017-04-11 10:38:48
function computed(money) {
var num = parseFloat(money) / 2;
var pingzi = num, gaizi = num;
var total = num;
reComputed();
function reComputed() {
if (pingzi < 2 && gaizi < 4) {
return;
}
if (pingzi >= 2) {
var beishu = Math.floor(pingzi / 2);
pingzi = pingzi - beishu * 2 + Math.floor(pingzi / 2);
gaizi = gaizi + beishu;
total = total + beishu;
} else {
var beishu_1 = Math.floor(gaizi / 4);
gaizi = gaizi - beishu_1 * 4 + Math.floor(gaizi / 4);
pingzi = pingzi + beishu_1;
total = total + beishu_1;
}
reComputed();
}
return total;
}输入金钱10，输出15，所有10元最多应该可以喝15瓶
伊谢尔伦2017-04-11 10:38:48
<!DOCTYPE html>
<html>
<head>
<title>测试</title>
<meta charset="utf-8">
</head>
<body>
<button id="testBtn">test</button>
<p style="width:300px;">
<p>啤酒2元一瓶,四个瓶盖可换一瓶啤酒,2个空瓶也可换一瓶啤酒,10元最多可以喝几瓶，要用js实现.下面输入所花费用</p>
<input type="text" name="fee" id="allfee">
<button id="openP">计算</button>
</p>
<script type="text/javascript">
window.onload = function() {
var allFee; //总费用
var count = 0; //当前已喝啤酒的数量
var pingzi = 0; //当前的瓶子数
var gaizi = 0; //当前的盖子数量
var calculateCount = document.getElementById("openP");
//点击开始计算
calculateCount.onclick = function(){
allFee = parseInt(document.getElementById('allfee').value)
step1(allFee);
}
//步骤1. 计算用当前费用,所能喝到的啤酒数量
function step1(fee){
let curpi1 = parseInt(fee/2);
count += curpi1;
pingzi += curpi1;
gaizi += curpi1;
step2();
}
//步骤2. 计算用瓶盖兑换啤酒喝的数量
function step2(){
if(gaizi/4 < 1 && pingzi/2 < 1){
alert(count);
return false;
}else{
let curpi2 = parseInt(gaizi/4);
count += curpi2;
pingzi += curpi2;
gaizi = gaizi%4 + curpi2;
step3();
}
}
//步骤3. 计算用空瓶子兑换啤酒喝的数量
function step3(){
if(gaizi/4 < 1 && pingzi/2 < 1){
alert(count);
return false;
}else{
let curpi3 = parseInt(pingzi/2);
count += curpi3;
gaizi += curpi3;
pingzi = pingzi%2 + curpi3;
step2();
}
}
}
</script>
</body>
</html>
高洛峰2017-04-11 10:38:48
function howMany(price, buttle, cap){/*兑换所需单价 瓶子 瓶盖*/
return function get(money, sum = 0, b = 0, pg = 0){
var n = money ? (money / price) << 1 >> 1 : ((b / buttle) << 1 >> 1) + ((pg / cap) << 1 >> 1);
return !n ? sum : get(0, sum + n, b % buttle + n, pg % cap + n);
}
}
var count = howMany(2, 2, 4);
count(10) 怀念pg作为某种货币的时光
PHPz2017-04-11 10:38:48
def beer(total, cover, bottle):
if cover < 4 and bottle < 2: return total
if cover >= 4: total, bottle, cover = total+cover/4, bottle+cover/4, cover%4+cover/4
if bottle >= 2: total, cover, bottle = total+bottle/2, cover+bottle/2, bottle%2+bottle/2
return beer(total, cover, bottle)print beer(10/2, 10/2, 10/2)
迷茫2017-04-11 10:38:48
感觉就得用递归啊
var fn=function(x,y,z,m){//x是完整瓶,y是空瓶,z是空盖,m是能喝的瓶数
return x==0&&y<2&&z<4?m:fn(parseInt(y/2)+parseInt(z/4),y%2+x,z%4+x,m+=x)}
var num=fn(10/2,0,0,0)