回答編集履歴
4
説明
test
CHANGED
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@@ -1,4 +1,8 @@
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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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+
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3
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+
引き直しなど条件分岐は行わず、
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4
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+
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5
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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
6
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3
7
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4
8
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3
修正
test
CHANGED
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@@ -26,14 +26,14 @@
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26
26
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27
27
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28
28
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29
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-
a
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29
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+
histogram=[0,0,0,0,0,0,0]
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30
30
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31
31
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for(i=0;i<1000;i++){
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32
32
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33
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-
a
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33
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+
histogram[g()+3] ++
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34
34
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35
35
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}
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36
36
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37
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-
console.log(a
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37
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+
console.log(histogram)
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38
38
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39
39
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```
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2
さらに改良
test
CHANGED
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@@ -12,6 +12,14 @@
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12
12
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13
13
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14
14
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15
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+
function g(){
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16
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+
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17
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+
r = Math.floor(Math.random() * 6)
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18
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+
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19
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+
return f(r)
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20
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+
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21
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+
}
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22
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+
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15
23
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16
24
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17
25
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console.log([0,1,2,3,4,5].map(f))
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@@ -22,9 +30,7 @@
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22
30
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23
31
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for(i=0;i<1000;i++){
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24
32
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25
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-
r = Math.floor(Math.random() * 6)
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26
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-
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27
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-
arr[
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33
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+
arr[g()+3] ++
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28
34
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29
35
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}
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30
36
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1
コード改良
test
CHANGED
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@@ -6,7 +6,7 @@
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6
6
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7
7
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function f(t){
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8
8
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9
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-
return Math.ceil(Math.abs(t-2.5))*
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9
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+
return Math.ceil(Math.abs(t-2.5))*Math.sign(t-2.5)
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10
10
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11
11
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}
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12
12
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@@ -14,18 +14,20 @@
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14
14
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15
15
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16
16
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17
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-
console.log([0,1,2,3,4,5].map(f))
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17
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+
console.log([0,1,2,3,4,5].map(f))
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18
18
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19
19
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20
20
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21
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+
arr=[0,0,0,0,0,0,0]
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21
22
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22
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-
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23
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-
for(i=0;i<
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23
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+
for(i=0;i<1000;i++){
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24
24
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25
25
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r = Math.floor(Math.random() * 6)
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26
26
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27
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-
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27
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+
arr[f(r)+3] ++
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28
28
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29
29
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}
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30
30
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31
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+
console.log(arr)
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32
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+
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31
33
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```
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