Chiaki Sequence Revisited題解

首先順着題目邏輯打表

打表代碼：

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int lowbit(int x)
{
	return x&(-x);
}
int main()
{
	int N;
	int a[1007] = { 1,1 };
	cin >> N;
	while (N--)
	{
		for (int n = 2; n < 100; ++n)
			a[n] = a[n - a[n - 1]] + a[n - 1 - a[n-2]];
		int sum = 0;
		for (int i = 0; i < 100; ++i)
		{
			sum += a[i];
			cout << "a" << i + 1 << " = " << a[i] << " " << "SUM";
			cout<< i + 1 << " = " << sum << "   " << "lowbit(" << i + 1 << ")  " << lowbit(i + 1) << endl;
		}
	}
	system("pause");
	return 0;
}

打表結果：

可以觀察出第n個數出現次數是lowbit(n).

《關於lowbit函數》https://blog.csdn.net/qq853674765/article/details/70050306--醉醉Philip

 

由此可得：

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int lowbit(int x)
{
	return x&(-x);
}
int main()
{
	int N;
	cin >> N;
	while (N--)
	{
		int n, j = 2;
		int k = 1;
		int sum;
		cin >> n;
		if (n > 2)
			sum = 2;
		else
			sum = n;
		for (int i = 2; i < n; ++i,++k)
		{
			sum += j;
			if (k == lowbit(j))
			{
				++j;
				k = 0;
			}
		}
		cout << sum << endl;
	}
	system("pause");
	return 0;
}