HDU3519 【矩阵快速幂】

As we all know,every coin has two sides,with one side facing up and another side facing down.Now,We consider two coins's state is same if they both facing up or down.If we have N coins and put them in a line,all of us know that it will be 2^N different ways.We call a "N coins sequence" as a Lucky Coins Sequence only if there exists more than two continuous coins's state are same.How many different Lucky Coins Sequences exist? Input There will be sevaral test cases.For each test case,the first line is only a positive integer n,which means n coins put in a line.Also,n not exceed 10^9. Output You should output the ways of lucky coins sequences exist with n coins ,but the answer will be very large,so you just output the answer module 10007. Sample Input 3 4 Sample Output 2 6

转载（看到一篇博客分析不错，就不自己写啦 ）

咋一看还以为是DP,当看到规模的时候傻眼了,10^9!!!!!这下咋办…

后来才知道是要推公式的,问了两位大牛,公式是这么推的:

长度为n的01串一共有2^n种不同的排列方法,

设f(n)为长度是n的不包含连续3个或以上相同的1或0的01串,

则f(1)=2,f(2)=4,f(3)=6,f(4)=10

当n>4的时候,分情况考虑:

1、  如果是以00或者11结尾,则分别有f(n-2)/2种情况,加起来就是f(n-2)种.

2、  如果是以01或者10结尾,则第n个字符要和第n-1个字符不一样,那么分别有f(n-1)/2种,加起来就是f(n-1)

则统计起来就是f(n)=f(n-1)+f(n-2),题目要求的是包含连续三个相同的0或1串的串数,那就是用a[n]=(2^n-f(n))007.

然而这样还不好求,先不看007,转换成递推公式是a[n]=a[n-1]+a[n-2]+2^(n-2),

转换成矩阵:

a[n]             1  1  1         a[n-1]

a[n-1]    =   1  0  0    *   a[n-2]

2^(n-1)       0  0  2         2^(n-2)

这样就可以用矩阵幂快速算出a[n],复杂度为O(logn)

 代码原创

#include<cstring> #include<cstdio> #include<iostream> using namespace std; const int N=10007; int x[4][4],ret[4][2],t[4][4]; void f1() { memset(t,0,sizeof(t)); for(int i=1;i<=3;i++) for(int k=1;k<=3;k++) t[i][1]=(t[i][1]+x[i][k]*ret[k][1])%N;; for(int i=1;i<=3;i++) ret[i][1]=t[i][1]; } void f2() { memset(t,0,sizeof(t)); for(int i=1;i<=3;i++) for(int k=1;k<=3;k++) for(int j=1;j<=3;j++) t[i][j]=(t[i][j]+x[i][k]*x[k][j])%N; for(int i=1;i<=3;i++) for(int j=1;j<=3;j++) x[i][j]=t[i][j]; } int main() { int a[5]={0,0,0,2,6}; int m; while(scanf("%d",&m)!=EOF) { if(m<=4){printf("%d\n",a[m]);continue;} m-=4; x[1][1]=x[1][2]=x[1][3]=x[2][1]=1; x[2][2]=x[2][3]=x[3][1]=x[3][2]=0; x[3][3]=2; ret[1][1]=6,ret[2][1]=2;ret[3][1]=8; while(m>0) { if(m&1)f1(); f2(); m=m/2; } printf("%d\n",ret[1][1]); } return 0; }