题解：CF1453D Checkpoints
提供一种直接基于期望推表达式的做法。
设 ppp 表示走一步成功的概率，E[x]E[x]E[x] 表示从一个存档点开始，还剩 xxx 步的距离到达下一个存档点，所需的期望步数。若一次成功，共走 xxx 步，概率为 pxp^xpx；若走到第 iii 步时失败，概率为 pi−1(1−p)p^{i-1}(1-p)pi−1(1−p)，之后需要重新开始，共走 E[x]E[x]E[x] 步。因此有表达式：
E[x]=x×px+∑i=1xpi−1(1−p)(i+E[x]) E[x] = x \times p^x + \sum \limits_{i = 1}^x p^{i - 1}(1-p)(i + E[x]) E[x]=x×px+i=1∑xpi−1(1−p)(i+E[x])
直接将 p=12p = \frac{1}{2}p=21 带入可知：
E[x]=x(12)x+2[1−(x+2)(12)x+1]+[1−(12)x]E[x](12)xE[x]=2−(12)x−1E[x]=2x+1−2 E[x] = x (\frac{1}{2})^x + 2[1 - (x + 2)(\frac{1}{2})^{x + 1}] + [1 - (\frac{1}{2})^x]E[x]\\ (\frac{1}{2})^xE[x] = 2 - (\frac{1}{2})^{x - 1}\\ E[x] = 2^{x + 1} - 2 E[x]=x(21)x+2[1−(x+2)(21)x+1]+[1−(21)x]E[x](21)xE[x]=2−(21)x−1E[x]=2x+1−2
因此直接构造 100...\texttt{100...}100... 的形式，类似二进制拆分进行构造即可。由于 E[x]E[x]E[x] 为偶数，所以奇数的情况显然无解。
代码如下：
#include <bits/stdc++.h>
#define pii pair <int,int>
#define init(x) memset (x,0,sizeof (x))
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 1e5 + 5;
const int MOD = 1e9 + 7;
inline ll read ();
void solve ()
{ll x = read ();if (x & 1) {puts ("-1");return;}vector <ll> f (61,0);for (int i = 0;i <= 60;++i) f[i] = (1ll << (i + 2)) - 2;vector <int> ans;for (int i = 60;~i;--i){while (x >= f[i]){x -= f[i];ans.push_back (1);for (int j = 1;j <= i;++j) ans.push_back (0);}}assert ((int)ans.size () <= 5000);printf ("%d\n",(int) ans.size ());for (auto v : ans) printf ("%d ",v);puts ("");
}
int main ()
{int t = read ();while (t--) solve ();return 0;
}
inline ll read ()
{ll s = 0;int f = 1;char ch = getchar ();while ((ch < '0' || ch > '9') && ch != EOF){if (ch == '-') f = -1;ch = getchar ();}while (ch >= '0' && ch <= '9'){s = s * 10 + ch - '0';ch = getchar ();}return s * f;
}