暴力枚举 ``` #include <cstdio> #include <cmath> using namespace std; int main(){ int n = 23333333; for(int i=1;i<n;i++) { double a = i *1.0/n; double b = (n-i)*1.0/n; double res =0; res-=a*log2(a)*i+b*log2(b)*(n-i); if(fabs(res-11625907.5798)<0.0001) { printf("%d\n",i); break; } } return 0; } ```

该死的公式可以化简为 $$ \frac{x^2}{n}log_2\frac{x}{n} + \frac{(n - x)^2}{n}log_2\frac{n-x}{n} $$ 其中 x 是 0 或者 1 在字符串中出现的次数, n - x 则是 1 或者 0 在字符串出现的次数, n 是字符串长度 然后枚举可能的出现次数即可, 注意控制精度🎉️ ```cpp #include <iostream> #include <cmath> using namespace std; int main() { int n = 23333333; int zf = n / 2; double s = 11625907.5798; // 遍历所有可能的 0 的数量 for (int i = 0; i < zf; ++i) { double x = n - i; double y = i; double jg = -((x*x)/n * log2(x/n) + ((y*y)/n * log2(y/n))); if (jg > s) { cout << jg << " ~ " << s << '\n'; if (s - jg < 1e-6) cout << i << '\n'; break; } } return 0; } ```

``` #include<iostream> #include<cstring> #include<algorithm> #include<math.h> using namespace std; #define N 23333333 int main() { for(int x=1;x<N/2;x++) { double p0=x*1.0/N,p1=(N-x)*1.0/N; if(abs(-x*p0*log2(p0)-(N-x)*p1*log2(p1)-11625907.5798)<0.0001) { cout<<x<<endl; return 0; } } } ```

本题主要是浮点数的比较一个需要知道的是fabs和abs区别，另一个是知道熵是什么，然后直接暴力就行，记住不要犯常规错误（如1e6）。 include //typedef long long ll; using namespace std; int main() { ``` int n=23333333; double check=11625907.5798; for(int z=1;z<=n;z++) { int one=n-z; if(z>one)break; double PZ=1.0\*z/n; double POne=1.0\*one/n; double HZ=-z\*PZ\*log2(PZ); double HOne=-one\*POne\*log2(POne); double HS=HZ+HOne; if(fabs(HS-check)<1e-4) { cout<<z; break; } } ``` }

``` #include <bits/stdc++.h> #define int long long using namespace std; signed main() { int n = 23333333 / 2; int m = n + 1; int q = 23333333; int i; for(i = n; i >= 0; i -- ) { double a0 = -1 * 1.0 * i * (1.0 * i / q) * log2(1.0 * i / q); double a1 = -1 * 1.0 * (q - i) * (1.0 * (q - i) / q) * log2(1.0 * (q - i) / q); double aa = a0 + a1; if(fabs(aa - 11625907.5798) < 0.0001) break; } cout << i; } ```

``` #include<bits/stdc++.h> using namespace std; const int N = 23333333; double num = 11625907.5798; double E = 1e-4; int ans; int main() { int l = 1, r = N / 2; while (l <= r) { int mid = (l + r) >> 1; int y = N - mid; double s = -1.0 * y * y / N * log2(1.0 * y / N) - 1.0 * mid * mid / N * log2(1.0 * mid / N); if (abs(s - num) <= E) { cout << mid; break; } else { if (s > num) r--; else if (s < num) l++; } } return 0; } ```

``` #include<iostream> #include<cmath> #include<algorithm> using namespace std; double res=11625907.5798; int sumNumber=23333333; double cul(int count0,int count1){ double bite0=count0*1.0/sumNumber,bite1=count1*1.0/sumNumber; double hs=-count0*(bite0*log2(bite0))-count1*(bite1*log2(bite1)); return hs; } int main(){ for(int i=1;i<=sumNumber/2;i++){ int count0=i,count1=sumNumber-i; if(abs(cul(count0,count1)-res)<1e-4){ cout<<count0; break; } } return 0; } ```

``` #include <bits/stdc++.h> using namespace std; const double N = 23333333; const string ans = "11625907.5798"; int main(void) { int t = 0; for (t = 0; t <= N; t += 1) { double p_0 = t / N; double p_1 = (N - t) / N; double tmp = ((t * -p_0 * log2(p_0)) + ((N - t) * -p_1 * log2(p_1))); char fin[ans.size()]; sprintf(fin, "%.4f", tmp); if (ans == fin) { cout << t; break; } } return 0; } ```

``` #include<bits/stdc++.h> #include<algorithm> #include<vector> using namespace std; typedef long double db; const int N = 23333333; const db ans = 11625907.5798, eps = 1e-4; int main() { ios::sync_with_stdio(0);cin.tie(0); for(int v = 0;v <= N / 2;++v) { int u = N - v; db res = -1.0 * u * u / N * log2(1.0 * u / N) - 1.0 * v * v / N * log2(1.0 * v / N); if(fabs(res - ans) < eps) { cout<<v<<endl; return 0; } } return 0; } ```