Problem of Precision（矩阵快速幂）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=2256

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1083    Accepted Submission(s): 645

Problem Description

 

Input

The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n. (1 <= n <= 10^9)

 

Output

For each input case, you should output the answer in one line.

 

Sample Input

3125

 

Sample Output

997841

 

Source

HDOJ 2008 Summer Exercise（4）- Buffet Dinner

 

题意：求(sqrt(2) + sqrt(3)) ^ 2n MOD 1024。

编程思想：参考以下盗来的推公式的图

AC code：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<queue>#include<map>#define LL long long#define MAXN 1000010using namespace std;const int INF=0x3f3f3f3f;//----以下为矩阵快速幂模板-----//const int mod=1024;const int NUM=11;//定义矩阵能表示的最大维数int N;//N表示矩阵的维数，以下的矩阵加法、乘法、快速幂都是按N维矩阵运算的struct Mat{//矩阵的类int a[NUM][NUM];void init()//将其初始化为单位矩阵{memset(a,0,sizeof(a));for(int i=0;i<NUM;i++){a[i][i]=1;}}};Mat add(Mat a,Mat b)//(a+b)%mod  矩阵加法{Mat ans;for(int i=0;i<N;i++){for(int j=0;j<N;j++){ans.a[i][j]=a.a[i][j]+b.a[i][j];ans.a[i][j]%=mod;}}return ans;}Mat mul(Mat a,Mat b) //(a*b)%mod  矩阵乘法{Mat ans;for(int i=0;i<N;i++){for(int j=0;j<N;j++){ans.a[i][j]=0;for(int k=0;k<N;k++){ans.a[i][j]+=a.a[i][k]*b.a[k][j];}ans.a[i][j]%=mod;}}return ans;}Mat power(Mat a,int num)//(a^n)%mod  矩阵快速幂{Mat ans;ans.init();while(num){if(num&1){ans=mul(ans,a);}num>>=1;a=mul(a,a);}return ans;}Mat pow_sum(Mat a,int num)//(a+a^2+a^3....+a^n)%mod 矩阵的幂和{int m;Mat ans,pre;if(num==1)return a;m=num/2;pre=pow_sum(a,m);ans=add(pre,mul(pre,power(a,m)));if(num&1)ans=add(ans,power(a,num));return ans;}void output(Mat a)//输出矩阵{for(int i=0;i<N;i++){for(int j=0;j<N;j++){printf("%d%c",a.a[i][j],j==N-1?'\n':' ');}}}//----以上为矩阵快速幂模板-----//int main(){int t,n,k,ans,i,j;int T;scanf("%d",&T);while(T--){    scanf("%d",&n);    if(n==1)        {            printf("9\n");            continue;        }Mat a,f;N=2;    a.a[0][0]=5;    a.a[0][1]=12;    a.a[1][0]=2;    a.a[1][1]=5;    k=n%1024;//k为循环节    if(k==0)            k=1024;f=power(a,k-1);int an=f.a[0][0]*5+f.a[0][1]*2;printf("%d\n",(2*an-1)%mod); }return 0;}