hdu4565(矩阵快速幂+经典的数学处理)

注意题目的一个关键条件(a-1)²< b < a^{2 ， 于是可以知道}    0 < a-√b < 1 ,所以 (a-√b)^n < 1 . 然后 (a+ √b)^n+(a-√b)^n 的值为整数且正好是 (a+√b)^n的向上取整.

然后就可以得到递推式 f[n+1]=2*a*f[n]-(a*a-b)*f[n-1] . 

然后构造矩阵。 幂乘即可. 

　　　　　　　　　　　　　　　　So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1364    Accepted Submission(s): 424

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

 

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

 

Recommend

zhoujiaqi2010

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#include <sstream>
#include <iostream>
using namespace std;
#define INF 0x3fffffff

typedef __int64 LL;

LL a,b,n,m;
LL g[2][2];
LL MOD;

void cal(LL s[2][2],LL t[2][2])
{
    LL tmp[2][2];
    memset(tmp,0,sizeof(tmp));
    for(int k=0;k<2;k++)
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++)
            {
                tmp[i][j]=(tmp[i][j]+s[i][k]*t[k][j])%MOD;
            }
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            s[i][j]=tmp[i][j];
}


int main()
{
    //freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
    //freopen("C:\Users\Administrator\Desktop\in.txt","w",stdout);
    while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF)
    {
        MOD=m;
        g[0][0]=2*a; g[0][1]=1;
        g[1][0]=b-a*a; g[1][1]=0;

        n--;
        LL sum[2][2];
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++)
                if(i==j) sum[i][j]=1;
                else sum[i][j]=0;
        while(n)
        {
            if( (n&1)!=0)
                cal(sum,g);
            cal(g,g);
            n>>=1;
        }
        LL ans=((2*a*sum[0][0]+2*sum[1][0])%MOD+MOD)%MOD;
        cout<<ans<<endl;
    }
    return 0;
}