回答編集履歴
2
再帰呼出しを使う解を追加
answer
CHANGED
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@@ -41,4 +41,38 @@
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41
41
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std::cout << " }\n";
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42
42
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}
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43
43
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}
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44
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+
```
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45
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+
**追記2**
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46
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再帰呼出しを使うと、
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47
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```C++
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48
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#include <iostream>
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49
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#include <algorithm> // copy
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50
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51
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const int N = 4; // N: 1, 2, 3, 4, 5, 6
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52
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const int M = 256; // NのN乗: 1, 4, 27, 256, 3125, 46656
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53
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int a[M][N], b[N], k = 0;
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54
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+
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55
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const int A = 1, B = 3, C = 4, D = 6, E = 7, F = 9;
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56
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const int n[] = { A, B, C, D, E, F };
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57
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+
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58
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void gen(int i)
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59
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{
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60
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if (i < N)
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61
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for (int j = 0; j < N; j++)
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62
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b[i] = n[j], gen(i + 1);
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63
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else
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64
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std::copy(b, b + N, a[k++]);
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65
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}
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66
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+
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67
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int main(void)
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68
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{
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69
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gen(0);
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70
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+
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71
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// 以下、確認
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72
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for (int i = 0; i < M; i++) {
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73
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std::cout << "a[" << i << "] = { " << a[i][0];
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74
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for (int j = 1; j < N; j++) std::cout << ", " << a[i][j];
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75
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std::cout << " }\n";
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76
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}
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77
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}
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44
78
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```
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1
別解の追加
answer
CHANGED
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@@ -15,4 +15,30 @@
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15
15
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std::cout << "a[" << i << "] = { " << a[i][0] << ", "
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16
16
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<< a[i][1] << ", " << a[i][2] << ", " << a[i][3] << " }\n";
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17
17
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}
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18
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```
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19
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+
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20
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**追記**
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21
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6通りの数値まで扱えるようにしてみました。
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22
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```C++
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23
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#include <iostream>
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24
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25
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const int N = 3; // N: 1, 2, 3, 4, 5, 6
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26
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const int M = 27; // NのN乗: 1, 4, 27, 256, 3125, 46656
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27
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int a[M][N];
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28
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+
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29
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int main()
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30
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{
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31
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int A = 1, B = 3, C = 4, D = 6, E = 7, F = 9;
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32
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int n[] = { A, B, C, D, E, F };
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33
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+
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34
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for (int i = 0; i < M; i++)
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35
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for (int v = i, j = N; --j >= 0; v /= N) a[i][j] = n[v % N];
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36
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+
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37
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// 以下、確認
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38
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for (int i = 0; i < M; i++) {
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39
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std::cout << "a[" << i << "] = { " << a[i][0];
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40
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for (int j = 1; j < N; j++) std::cout << ", " << a[i][j];
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41
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std::cout << " }\n";
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42
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}
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43
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}
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18
44
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```
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