HDU 4565 So Easy!(矩阵快速幂)

这里写的很详细了啊：http://blog.csdn.net/acmmmm/article/details/9722515?utm_source=tuicool

记录一下，回头复习用啊。

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2281    Accepted Submission(s): 708

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 20132 3 2 20132 2 1 2013

 

Sample Output

4144

 

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

#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 10007const 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;}