Description
剛開始你有一個數字0，每一秒鐘你會隨機選擇一個[0,2^n-1]的數字，與你手上的數字進行或（c++,c的|,pascal 的or）操作。選擇數字i的概率是p[i]。保證0<=p[i]<=1，Σp[i]=1問期望多少秒后，你手上的數字變成2^n-1。 Input
第一行輸入n表示n個元素，第二行輸入2^n個數，第i個數表示選到i-1的概率
Output
僅輸出一個數表示答案，絕對誤差或相對誤差不超過1e-6即可算通過。如果無解則要輸出INF
Sample Input
2
0.250.250.250.25 Sample Output
2.6666666667 HINT
對于100%的數據，n<=20
Source
鳴謝bhiaibogf提供試題，istream提供SPJ
論文題 http://ydc.blog.uoj.ac/blog/336
#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=PRe[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld/n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { / For(j,m-1) cout<<a[i][j]<<' ';/ cout<<a[i][m]<<endl; / } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){ int x=0,f=1; char ch=getchar(); while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();} while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();} return x*f;} #define MAXN (1<<21)int n;double a[MAXN];int main(){// freopen("bzoj4036.in","r",stdin);// freopen(".out","w",stdout); cin>>n; int S=1<<n; Rep(i,S) cin>>a[i]; Rep(i,n) Rep(j,S) if (j>>i&1) a[j]+=a[j^(1<<i)]; Rep(i,S) { if (a[i]>=1-1e-8) { if (i==S-1) a[i]=0; else {puts("INF"); return 0;} } else a[i]=-1./(1-a[i]); } Rep(i,n) Rep(j,S) if (j>>i&1) a[j]-=a[j^(1<<i)]; printf("%.10lf/n",a[S-1]); return 0;}新聞熱點
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