hdu 4565

矩阵快速幂

1、http://acm.hdu.edu.cn/showproblem.php?pid=4565

2、

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 187    Accepted Submission(s): 67

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

 

 

Recommend

zhoujiaqi2010

 

3、代码：

#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream>#include<math.h>using namespace std;#define ll long longint aa,b,n,m;struct matrix{    ll a[2][2];} origin,res;matrix multiply(matrix x,matrix y){    matrix temp;    memset(temp.a,0,sizeof(temp.a));    for(int i=0; i<2; i++)    {        for(int j=0; j<2; j++)        {            for(int k=0; k<2; k++)            {                temp.a[i][j]+=(((x.a[i][k])%m)*((y.a[k][j])%m))%m;            }            temp.a[i][j]%=m;        }    }    return temp;}void init(){    origin.a[0][0]=(2*(aa%m))%m;    origin.a[0][1]=(b%m-((aa%m)*(aa%m))%m+m)%m;    origin.a[1][0]=1;    origin.a[1][1]=0;    memset(res.a,0,sizeof(res.a));    res.a[0][0]=res.a[1][1]=1;}void calc(int N){    int n=N;    while(n)    {        if(n&1)            res=multiply(res,origin);        n>>=1;        origin=multiply(origin,origin);    }    matrix tmp,tmp2;    tmp2.a[0][0]=(((2*(aa%m)*(aa%m))%m)%m+2*(b%m))%m;    tmp2.a[0][1]=0;    tmp2.a[1][0]=(2*(aa%m))%m;    tmp2.a[1][1]=0;    if(N==0)        printf("%lld\n",tmp2.a[0][0]%m);    else    {        tmp=multiply(res,tmp2);        printf("%lld\n",tmp.a[0][0]%m);    }}int main(){    while(cin>>aa>>b>>n>>m)    {        if(n==1)        {            printf("%lld\n", (ll)ceil((aa*1.0+sqrt(b*1.0))*1.0)%m);        }        else        {            init();            calc(n-2);        }    }    return 0;}/*2 3 10 20132 3 2 20132 2 1 2013*/