每次都和e[0]有关系 通过方程消去环

dp[i] = sigma(dp[i+k]*p)+dp[0]*p+1
dp[i] = a[i]*dp[0]+b[i]
dp[i] = sigma(p*(a[i+k]*dp[0]+b[i+k]))+dp[0]*p+1
a[i] = sigma(a[i+k]*p)+p
b[i] = sigma(b[i+k]*p)+1

#include 
  
   
#include 
   
     using namespace std; double A[555], B[555], P[555]; //dp[i] = sigma(dp[i+k]*p)+dp[0]*p+1 //dp[i] = a[i]*dp[0]+b[i] //dp[i] = sigma(p*(a[i+k]*dp[0]+b[i+k]))+dp[0]*p+1 //a[i] = sigma(a[i+k]*p)+p //b[i] = sigma(b[i+k]*p)+1 int main() { int T; scanf("%d", &T); while(T--) { int n, k1, k2, k3, a, b, c; scanf("%d %d %d %d %d %d %d", &n, &k1, &k2, &k3, &a, &b, &c); memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B)); memset(P, 0, sizeof(P)); double p = 1.0/(k1*k2*k3); for(int i = 1; i <= k1; i++) for(int j = 1; j <= k2; j++) for(int k = 1; k <= k3; k++) if(i != a || j != b || k != c) P[i+j+k] += p; for(int i = n; i >= 0; i--) { A[i] = p; B[i] = 1; for(int j = 1; j <= k1+k2+k3; j++) { A[i] += A[i+j]*P[j]; B[i] += B[i+j]*P[j]; } } printf("%.18lf\n", B[0]/(1-A[0])); } return 0; }