回答編集履歴
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追加。
test
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@@ -5,3 +5,195 @@
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使える数字に’０’があった場合、頭が’０’のものは’０’だけである。（解の１つ）
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後は桁数になる数字の組み合わせをカウントして、その組み合わせ数が前述の’０’が選択できる数字にある場合は＋１して偶数か奇数かを判断し、それを算出する。
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15桁までという事を考えてlong long intで処理してみる。
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```C++
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#include <iostream>
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#include <string>
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#include <vector>
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unsigned char used_number[10]; // 使用数字 0=not used,1=used
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long long int check_number(int digit)
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{
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long long int work;
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if( digit == 1 )
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{
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for(int i = 0,work = 0LL;i <10;i++)
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if(used_number[i] != 0)
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work++;
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return work;
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}
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for(int i = 0;i<10;i++)
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work += check_number(digit-1);
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if(used_number[0] == 1) // 頭が’0’場合、その数を引く
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work -= number_of_count;
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return work;
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}
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long long int calc_number(int digit,long long int current_pos,long long int result_pos)
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{
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long long work;
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if( digit == 1)
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{
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for(int i = 0,work = 0LL;i<10;i++ )
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{
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if(used_number[i] != 0 )
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{
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if(work == result_pos )
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{
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cout << used_number[i] ;
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return;
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}
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}
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}
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return work;
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}
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for(int i = 0;i<10;i++)
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{
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current_pos += calc_number(digit-1,current_pos,result_pos);
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}
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if(used_number[0] == 1) // 頭が’0’場合、その数を引く
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current_pos -= number_of_count;
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return current_pos;
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}
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int main(void) {
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int digit;
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std::string number;
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long long int work;
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int number_count;
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long long int count_of_result=0LL;
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std::cin >> digit >> number；
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if((digit <= 0) || ( digit >15))
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{
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cout << "Illegal number" << endl;
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return -1;
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}
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// 重複確認
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for(int i = 0;i<10;i++)
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used_number[i]= 0;
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for(int i = 0,number_count =0;i<strlen(number);i++)
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{
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if((number[i] >= '0')&&(number[i] <='9'))
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{
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if( used_number[(number[i]-'0')] == 0 )
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{
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number_of_count++;
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used_number[(number[i]-'0')]=1;
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}
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}
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else {
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cout << "Illegal number" << endl;
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return -1;
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}
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}
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// 解の数を算出
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work = check_number(ditit);
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//
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if( work & 1 ) //解が奇数
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{ // 答えは２つ
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cout << calc_number(work/2) << "," << calc_number((work/2)+1) << endl;
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} else
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{ // 答えは１つ
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cout << calc_number(work/2) << endl;
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}
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}
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```
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机上なのでバグあるだろうなぁ。
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