Description
刚开始你有一个数字0，每一秒钟你会随机选择一个[0,2^n-1]的数字，与你手上的数字进行或（c++,c的|,pascal 的or）操作。选择数字i的概率是p[i]。保证0<=p[i]<=1，Σp[i]=1问期望多少秒后，你手上的数字变成2^n-1。 Input
第一行输入n表示n个元素，第二行输入2^n个数，第i个数表示选到i-1的概率
Output
仅输出一个数表示答案，绝对误差或相对误差不超过1e-6即可算通过。如果无解则要输出INF
Sample Input
2
0.250.250.250.25 Sample Output
2.6666666667 HINT
对于100%的数据，n<=20
Source
鸣谢bhiaibogf提供试题，istream提供SPJ
论文题 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;}新闻热点
疑难解答
