回答編集履歴
4
説明
answer
CHANGED
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@@ -1,4 +1,6 @@
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1
1
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無理やりやってみました。
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2
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+
引き直しなど条件分岐は行わず、
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3
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+
[0,5]の整数の一様分布に対してちゃんと均等な出現確率になるようになっています(多分)
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2
4
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3
5
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```javascript
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4
6
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function f(t){
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3
修正
answer
CHANGED
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@@ -12,9 +12,9 @@
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12
12
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13
13
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console.log([0,1,2,3,4,5].map(f))
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14
14
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15
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-
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15
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+
histogram=[0,0,0,0,0,0,0]
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16
16
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for(i=0;i<1000;i++){
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17
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-
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17
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+
histogram[g()+3] ++
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18
18
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}
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19
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-
console.log(
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19
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+
console.log(histogram)
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20
20
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```
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2
さらに改良
answer
CHANGED
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@@ -5,13 +5,16 @@
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5
5
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return Math.ceil(Math.abs(t-2.5))*Math.sign(t-2.5)
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6
6
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}
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7
7
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8
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+
function g(){
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9
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+
r = Math.floor(Math.random() * 6)
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10
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+
return f(r)
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11
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+
}
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8
12
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9
13
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console.log([0,1,2,3,4,5].map(f))
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10
14
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11
15
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arr=[0,0,0,0,0,0,0]
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12
16
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for(i=0;i<1000;i++){
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13
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-
r = Math.floor(Math.random() * 6)
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14
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-
arr[
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17
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+
arr[g()+3] ++
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15
18
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}
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16
19
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console.log(arr)
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17
20
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```
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1
コード改良
answer
CHANGED
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@@ -2,15 +2,16 @@
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2
2
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3
3
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```javascript
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4
4
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function f(t){
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5
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-
return Math.ceil(Math.abs(t-2.5))*
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5
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+
return Math.ceil(Math.abs(t-2.5))*Math.sign(t-2.5)
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6
6
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}
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7
7
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8
8
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9
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-
console.log([0,1,2,3,4,5].map(f))
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9
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+
console.log([0,1,2,3,4,5].map(f))
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10
10
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11
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-
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11
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+
arr=[0,0,0,0,0,0,0]
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12
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-
for(i=0;i<
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12
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+
for(i=0;i<1000;i++){
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13
13
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r = Math.floor(Math.random() * 6)
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14
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-
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14
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+
arr[f(r)+3] ++
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15
15
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}
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16
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+
console.log(arr)
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16
17
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```
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