PokemonUltraSun(评测)

Two pokemons are in a battle.One is our and another is the opposite’s.

Our pokemon is in confusion and the opposite’s pokemon is frozen.

Once per turn , the opposite’s pokemon does nothing and our pokemon gives w damages to the opposite’s pokemon with a probability of P while gives w damages to itself with a probability of 1 − P.

Our pokemon has hp1 health points and the opposite’s pokemon has hp2 health points.

If one pokemon’s health points are zero or below zero, the game is over.

Your task is to calculate the expected number of turn.

The ﬁrst line is an integer T(T ≤ 8) , the number of test cases.

For each test case, the ﬁrst line contains three integers hp1, hp2, w(1 ≤ hp1, hp2, w ≤ 3000), representing our pokemon’s health points, the opposite’s health points and damage. The second line contains a real number P(0 < P < 1). which is described above.

#include<bits/stdc++.h> using namespace std; #define f(i,b) for(int i=1;i<=b;++i) double d[3005][3005],P;int main(){int t,H,h,w;cin>>t; while(t--){cin>>H>>h>>w>>P;int a=(H+w-1)/w,b=(h+w-1)/w; f(i,a)f(j,b)d[i][j]=d[i-1][j]*(1-P)+d[i][j-1]*P+1; printf("%.6lf\n",d[a][b]); }return 0;}

详情