HDU 4870 rating (probability, expectation, push formula) & zoj 3415 zhou Yu

In fact, zoj 3415 should be called Yu Zhou... After encountering zoj 3415, I used the second method of Reference URL to seek the communication item, and this time I met 4870, I won't do it. After referring to chanme, I re-deduced zhou Yu and the ranking repeatedly (not to push down) four or five times to write this question about plagiarism...

These two questions are very similar. Multi-school rating is equivalent to a forced version, but the principles are the same. It seems that Gaussian yuan can be used for elimination, but I won't. The formula is pushed silently.

Formula Derivation reference http://www.cnblogs.com/chanme/p/3861766.html#2993306

Http://www.cnblogs.com/lijunle/archive/2010/10/09/1846577.html

They are different. Now I feel better at understanding them.

 

HDU 4870 rating

First, consider registering only one account (for the initial E [0], that is, the expectation of 0 to 20, with E [20] = 0)

E [0] = p * E [1] + (1-p) * E [0] + 1 => E [0] = E [1] + 1/P

E [1] = p * E [2] + (1-p) * E [0] + 1 ==> E [0] = E [2] + 1/p + 1/(p * P)

E [2] = p * E [3] + (1-p) * E [0] + 1

E [n-1] = p * E [N] + (1-p) * E [n-3] + 1

E [N] = 0

Obviously, E [0] = E [k] + T [K]. (Because after each input, E [0] and e [k] will be multiplied by the upper coefficient p)

In general, E [k] = p * E [k + 1] + (1-p) * E [K-2] + 1, -T [k] = p * (-T [k + 1]) + (1-p) * (-T [K-2]) + 1

So with T [0] = 0, T [1] = 1/P, T [2] = 1/(p * P) and T [K], T [k + 1], t [K-2], we can find all t [I]

When there are two accounts, this is the result (0, 0) => (1, 0) =>) => .... => (19,19) => (20,19)

T [k] indicates the expected time for an account to reach K from 0, so t [20] + T [19] is the answer.

 

 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5  6 int main(){ 7     double p,t[22]; 8     while(~scanf("%lf",&p)){ 9         t[0]=0,t[1]=1./p,t[2]=1./p+1./(p*p);10         for(int i=3;i<=20;++i) t[i] = (1./p+1./p*t[i-1]-(1-p)/p*t[i-3]);11         printf("%.6lf\n",t[19]+t[20]);12     }13     return 0;14 }

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Although I think the reference is very, very ...... Very, very, very similar...

 

Zhou Yu, zoj 3415

(The initial E [N], that is, the expectation of N to 0, with E [0] = 0)

Similar Method, do not want to play the derivation process...

 1 #include <cstdio> 2 #include <iostream> 3 using namespace std; 4  5 double pp(double a,int n){ 6     double ret=1; 7     while(n){ 8         if(n&1) ret *= a; 9         a*=a;10         n>>=1;11     }12     return ret;13 }14  int main(){15      int n,m;16      while(~scanf("%d%d",&n,&m)){17          if(m==2){printf("%.10lf\n",1.*n*(n+1));continue;}18          double ans = 1.*m/(m-2)/(m-2);19          double tmp = pp(1./(m-1),n) + 1.*n*(m-2) - 1;20          printf("%.10lf\n",ans*tmp);21      }22      return 0;23  }

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