HDU 4565 So Easy!(公式化简+矩阵)

转载：http://www.klogk.com/posts/hdu4565/

这里写的非常好，看看就知道了啊。

题意很easy。a,b,n都是正整数。求

Sn=⌈(a+b√)n⌉%m,(a−1)2<b<a2

这个题目也是2008年Google Codejam Round 1A的C题。

做法事实上很easy。记(a+b√)n为An，配项

Cn=An+Bn=(a+b√)n+(a−b√)n

两项恰好共轭，所以Cn是整数。

又依据限制条件

(a−1)2<b<a2⇒0<a−b√<1⇒0<(a−b√)n<1⇒Bn<1

也就是说Cn=⌈An⌉

Sn=(Cn)%m

求Cn的方法是递推。
对Cn乘以(a+b√)+(a−b√)

Cn[(a+b√)+(a−b√)]=[(a+b√)n+(a−b√)n][(a+b√)+(a−b√)]=(a+b√)n+1+(a−b√)n+1+(a+b√)n(a−b√)+(a−b√)n(a+b√)=Cn+1+(a2−b)(a+b√)n−1+(a2−b)(a−b√)n−1=Cn+1+(a2−b)Cn−1

于是

Cn+1=2aCn−(a2−b)Cn−1

把这个递推式写成矩阵形式

[Cn+1Cn]=[2a1−(a2−b)0][CnCn−1]

于是就能够用矩阵高速幂来做了

[Cn+1Cn]=[2a1−(a2−b)0]n[C1C0]

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2286    Accepted Submission(s): 710

Problem Description

　　A sequence S_n is defined as:




Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S_n.

　　You, a top coder, say: So easy! 

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2¹⁵, (a-1)²< b < a², 0 < b, n < 2³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S_n.

Sample Input

2 3 1 2013
2 3 2 2013
2 2 1 2013

Sample Output

4
14
4

<span style="font-size:18px;">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

///#define mod 10007

const int maxn = 210;

using namespace std;

struct matrix
{
    LL f[3][3];
};
LL mod;

matrix mul(matrix a, matrix b, int n)
{
    matrix c;
    memset(c.f, 0, sizeof(c.f));
    for(int k = 0;k < n; k++)
    {
        for(int i = 0; i < n;i++)
        {
            if(!a.f[i][k]) continue;
            for(int j = 0; j < n; j++)
            {
                if(!b.f[k][j]) continue;
                c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j]+mod)%mod;
            }
        }
    }
    return c;
}
matrix pow_mod(matrix a, LL b, int n)
{
    matrix s;
    memset(s.f, 0 , sizeof(s.f));
    for(int i = 0; i < n; i++) s.f[i][i] = 1LL;
    while(b)
    {
        if(b&1) s = mul(s, a, n);
        a = mul(a, a, n);
        b >>= 1;
    }
    return s;
}

int main()
{
    LL a, b, n;
    while(~scanf("%I64d %I64d %I64d %I64d",&a, &b, &n, &mod))
    {
        if(n == 1)
        {
            printf("%I64d\n",2*a%mod);
            continue;
        }
        matrix c;
        memset(c.f, 0 , sizeof(c.f));
        c.f[0][0] = 2LL*a;
        c.f[0][1] = b-a*a;
        c.f[1][0] = 1LL;
        matrix d = pow_mod(c, n-1, 2);
        printf("%I64d\n",((d.f[0][0]*2*a+d.f[0][1]*2)%mod+mod)%mod);
    }
    return 0;
}</span>