bzoj4008[HNOI2015]亚瑟王
http://www.lydsy.com/JudgeOnline/problem.php?id=4008
我们只需要求每张牌发动的概率$P[i]$,然后乘上每张牌的伤害值$d[i]$即可。
记$f[i][j]$表示,在这$r$轮游戏中,有$j$轮游戏在第$i$张牌或第$i$张牌之前已经结束的概率。
那么还有$r-j$轮游戏在第$i$张牌时还没结束。
考虑第$i+1$张牌
如果这剩下的$r-j$轮游戏中,第$i+1$张牌都没有发动:
概率是$(1-p[i+1])^{r-j}$
转移到$f[i+1][j]$:
$f[i+1][j]+=f[i][j]*(1-p[i+1])^{r-j}$
如果这剩下的$r-j$轮游戏中,第$i+1$张牌在其中一局中发动了:
概率是$1-(1-p[i+1])^{r-j}$
转移到$f[i+1][j+1]$:
$f[i+1][j+1]+=f[i][j]*\{1-(1-p[i+1])^{r-j}\}$
这时候第$i+1$张牌的概率$p[i+1]$加上$f[i][j]*\{1-(1-p[i+1])^{r-j}\}$。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj using namespace std; typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP; #define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define fill(a,l,r,v) fill(a+l,a+r+1,v) #define re(i,a,b) for(i=(a);i<=(b);i++) #define red(i,a,b) for(i=(a);i>=(b);i--) #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define p_b(a) push_back(a) #define SF scanf #define PF printf #define two(k) (1<<(k)) template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;} inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0); int gint() { int res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z=='-'){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar()); return (neg)?-res:res; } LL gll() { LL res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z=='-'){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar()); return (neg)?-res:res; } const int maxn=220; const int maxr=132; int n,r; DB p[maxn+10],d[maxn+10]; DB f[maxn+10][maxr+10],ans; DB power(DB a,int k){DB x=1.0;while(k){if(k&1)x*=a;a*=a;k>>=1;}return x;} int main() { freopen("bzoj4008.in","r",stdin); freopen("bzoj4008.out","w",stdout); int i,j; for(int Case=gint();Case;Case--) { n=gint();r=gint(); re(i,1,n)scanf("%lf%lf\n",&p[i],&d[i]); re(i,0,maxn+10-1)re(j,0,maxr+10-1)f[i][j]=0; ans=0; f[0][0]=1; re(i,0,n-1) re(j,0,min(i,r)) { DB tmp=power(1-p[i+1],r-j); f[i+1][j]+=f[i][j]*tmp; if(j+1<=r) { f[i+1][j+1]+=f[i][j]*(1-tmp); ans+=f[i][j]*(1-tmp)*d[i+1]; } } PF("%0.10lf\n",ans); } return 0; }