HDU 2256 problem of precision (matrix fast power)

Address: HDU 2256

Ideas:

(SQRT (2) + SQRT (3) ^ 2 * n = (5 + 2 * SQRT (6) ^ N;

Note that (5 + 2 * SQRT (6) ^ N can be expressed as an + BN * SQRT (6 );

An + BN * (SQRT (6) = (5 + 2 * SQRT (6) * (A (n-1) + B (n-1) * SQRT (6 ))

= (5 * a (n-1) + 12 * B (n-1) + (2 * a (n-1) + 5 * B (n-1) * SQRT (6 );

Obviously, an = 5 * a (n-1) + 12 * B (n-1); bn = 2 * a (n-1) + 5 * B (n-1 );

At this time, it is easy to construct a matrix to quickly obtain an and BN:

5, 12

2, 5

What should we do next? The best way for me to wait is, of course .. Create a table .. Find the rule ..

Then the rule is ans = 2 * an-1;

How can we prove it? The proof is as follows:

(5 + 2 * SQRT (6) ^ n = An + BN * SQRT (6); (5-2 * SQRT (6 )) ^ n = An-BN * SQRT (6 );

(5 + 2 * SQRT (6) ^ N + (5-2 * SQRT (6) ^ n = 2 *;

Then, because

(5-2 * SQRT (6) ^ n = (0. 101...) ^ n <1;

Because

(5 + 2 * SQRT (6) ^ n = 2 * an-(5-2 * SQRT (6) ^ n

Available

2 * an-1 <(5 + 2 * SQRT (6) ^ n <2 *;

Therefore, the result of rounding (5 + 2 * SQRT (6) ^ n down must be 2 * an-1;

The proof is complete.

Therefore, you only need to use the Matrix to quickly obtain an power.

The Code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;const int mod=1024;struct matrix{    int ma[3][3];}init, res;matrix Mult(matrix x, matrix y){    matrix tmp;    int i, j, k;    for(i=0;i<2;i++)    {        for(j=0;j<2;j++)        {            tmp.ma[i][j]=0;            for(k=0;k<2;k++)            {                tmp.ma[i][j]=(tmp.ma[i][j]+x.ma[i][k]*y.ma[k][j])%mod;            }        }    }    return tmp;}matrix Pow(matrix x, int k){    int i, j;    matrix tmp;    for(i=0;i<2;i++) for(j=0;j<2;j++) tmp.ma[i][j]=(i==j);    while(k)    {        if(k&1) tmp=Mult(tmp,x);        x=Mult(x,x);        k>>=1;    }    return tmp;}int main(){    int t, k;    scanf("%d",&t);    while(t--)    {        scanf("%d",&k);        init.ma[0][0]=5;        init.ma[0][1]=12;        init.ma[1][0]=2;        init.ma[1][1]=5;        res=Pow(init,k-1);        int ans=(2*(res.ma[0][0]*5+res.ma[0][1]*2)-1)%mod;        printf("%d\n",ans);    }    return 0;}

HDU 2256 problem of precision (matrix fast power)