hdu-2256(矩阵快速幂+推导)

问题描述:

一句话题目:让你求(sqrt(2)+sqrt(3))^(2*n) mod 1024.

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

题目分析:

先给出一个大佬的博客，自己太鶸了，这个公式和一开始的想法完全不一样，后来发现我最初的想法是错的（千万别想着用double 取模)

点击打开链接

然后就可以根据那个矩阵来写代码了。

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;const int mod=1024;struct matrix//构建矩阵{    int f[3][3];    matrix operator* (const matrix &a) const {        matrix res;        for (int i=1;i<=2;i++) {            for (int j=1;j<=2;j++) {                res.f[i][j]=0;                for (int k=1;k<=2;k++)                    res.f[i][j]=(res.f[i][j]+(*this).f[i][k]*a.f[k][j])%mod;            }        }        return res;    }}a,b;void init()//预处理矩阵{    b.f[1][1]=5,b.f[1][2]=12;    b.f[2][1]=2,b.f[2][2]=5;    a.f[1][1]=5;    a.f[2][1]=2;}matrix fast_pow(matrix base,int k)//快速幂{    matrix ans=base;    while (k) {        if (k&1)            ans=base*ans;        base=base*base;        k>>=1;    }    return ans;}int main(){    int t;    scanf("%d",&t);    while (t--) {        int n;        scanf("%d",&n);        if (n==1) { printf("%d\n",9); continue;}        n-=2;        init();        struct matrix cur,ans;        cur=b;        cur=fast_pow(cur,n);        ans=cur*a;        printf("%d\n",(2*ans.f[1][1]-1)%mod);    }    return 0;}