hdoj.2256 Problem of Precision【矩阵快速幂】 2015/08/18

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1043    Accepted Submission(s): 612

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 

注：(√2+√3)^2n = ((√2+√3)^2)^n = (5+2√6)^n

由于√6消除不掉，设(5+2√6)^n = an+bn√6

则第n+1项 a(n+1) + b(n+1)√6 = (5+2√6)*(an + bn√6) = (5an+12bn) + (2an+5bn)√6

即a(n+1) = 5an+12bn,，b(n+1) = 2an+5bn

可由此构建矩阵

| 5 12 | *| an | = | a(n+1) |

| 2 5   |   | bn |    | b(n+1) |

再 (5+2√6)^n 向下取整   首先double不可取余操作

故令 (5+2√6)^n + (5-2√6)^n = (an+bn√6) + (an-bn√6) = 2an

又(5-2√6) ≈ 0.101020...  (5-2√6)^n 无限趋近于 0   

所以(5+2√6)^n 向下取整   即为 2an-1

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct node{    int m[2][2];}res,per;int t,n;const int m = 1024;void init(){    per.m[0][0] = per.m[1][1] = 1;    per.m[0][1] = per.m[1][0] = 0;    res.m[0][0] = res.m[1][1] = 5;    res.m[0][1] = 12;    res.m[1][0] = 2;}node mutil(node a,node b){    int i,j,k;    node c;    for( i = 0 ; i< 2 ; ++i )        for( j = 0 ; j < 2 ; ++j ){            c.m[i][j] = 0;            for( k = 0 ; k < 2 ; ++k )                c.m[i][j] += (a.m[i][k] * b.m[k][j])%m;            c.m[i][j] %= m;        }    return c;}void modefy( int k ){    node ans = per,p = res;    while( k ){        if( k&1 ) ans = mutil(ans,p);        p = mutil(p,p);        k>>=1;    }//    for( int i = 0 ; i < 2 ; ++i )//        printf("----  %d  %d\n",ans.m[i][0],ans.m[i][1]);    printf("%d\n",(10*ans.m[0][0] + 4*ans.m[0][1] - 1) % m);}int main(){    init();    scanf("%d",&t);    while(t--){        scanf("%d",&n);        if( n == 1 )            printf("9\n");        else modefy(n-1);    }    return 0;}