codeforces 717 D. Dexterina’s Lab(矩阵快速幂加速dp,好题)

题目链接

D. Dexterina’s Lab

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Dexterina and Womandark have been arch-rivals since they’ve known each other. Since both are super-intelligent teenage girls, they’ve always been trying to solve their disputes in a peaceful and nonviolent way. After god knows how many different challenges they’ve given to one another, their score is equal and they’re both desperately trying to best the other in various games of wits. This time, Dexterina challenged Womandark to a game of Nim.

Nim is a two-player game in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects from a single heap. The player who can't make a turn loses. By their agreement, the sizes of piles are selected randomly from the range [0, x]. Each pile's size is taken independently from the same probability distribution that is known before the start of the game.

Womandark is coming up with a brand new and evil idea on how to thwart Dexterina’s plans, so she hasn’t got much spare time. She, however, offered you some tips on looking fabulous in exchange for helping her win in Nim. Your task is to tell her what is the probability that the first player to play wins, given the rules as above.

Input

The first line of the input contains two integers n (1 ≤ n ≤ 109) and x (1 ≤ x ≤ 100) — the number of heaps and the maximum number of objects in a heap, respectively. The second line contains x + 1 real numbers, given with up to 6 decimal places each: P(0), P(1), ... , P(X). Here, P(i) is the probability of a heap having exactly i objects in start of a game. It's guaranteed that the sum of all P(i) is equal to 1.

Output

Output a single real number, the probability that the first player wins. The answer will be judged as correct if it differs from the correct answer by at most 10 - 6.

Example

input

2 20.500000 0.250000 0.250000

output

0.62500000

题意：

给出n，x，n (1 ≤ n ≤ 109) and x (1 ≤ x ≤ 100)，表示一共n堆石子，没堆石子的个数在[1,x]范围内，然后给出x+1个浮点数，分别表示没堆石子个数为0，1，2...x的概率，两人玩Nim游戏，求先手必胜的概率。

题解：

Nim游戏先手必胜条件为每堆石子个数xor和不为0.

注意到x的范围比较小 x (1 ≤ x ≤ 100)，可以记录当i堆石子确定时xor和为x的情况，那么

d[i][x]=d[i-1][x]*p0+d[i-1][x^1]*p1+...+d[i-1][x^n]*pn;

由于n比较大，可以用矩阵快速幂优化。

矩阵的构造自己画一下就知道了。由于xor和不会超过127，可以构造一个128*128的矩阵（0-127）。

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;const int N=130;struct Matrix{    int n;    double a[N][N];    Matrix(int in=0,int f=0)    {        n=in; for (int i=0;i<=n;i++) for (int j=0;j<=n;j++) a[i][j]=0;        if (f==1) for (int i=0;i<=n;i++) a[i][i]=1;    }    double *operator [](int x)    {        return a[x];    }    friend Matrix operator * (Matrix &A,Matrix &B)    {        int n=A.n; Matrix ret(n);        for (int i=0;i<=n;i++)            for (int j=0;j<=n;j++)                for (int k=0;k<=n;k++)                    ret[i][j]+=A[i][k]*B[k][j];        return ret;    }    friend Matrix Pow(Matrix &A,int b)    {        Matrix ret(A.n,1);        for (;b;b>>=1,A=A*A)            if (b&1)                ret=ret*A;        return ret;    }}A;int n,m;double a[1005];int main(){    scanf("%d%d",&n,&m);    for (int i=0;i<=m;i++) scanf("%lf",a+i);    A.n=127;    for (int i=0;i<=127;i++)        for (int j=0;j<=127;j++)            A[j][j^i]=a[i];    A=Pow(A,n);    printf("%.8lf\n",1-A[0][0]);}