HDU 4565 So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3459    Accepted Submission(s): 1113

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

 

 

 

解析：矩阵快速幂。

根据共轭式的原理，(a+√b)ⁿ+(a-√b)ⁿ必为整数。

而(a-1)²< b < a²

=> a-1< √b < a

=> 0< a-√b <1

=> 0< (a-√b)ⁿ <1

所以(a+√b)ⁿ+(a-√b)ⁿ = ceil( (a+√b)ⁿ )。

接下来就可以运用矩阵快速幂求解。

 

 

 

#include <cstdio>
#include <cstring>

long long a,b,n,m;

struct Mat1{
    long long mat[2][2];
};

struct Mat2{
    long long mat[2][1];
};

Mat1 operator * (Mat1 A,Mat1 B)
{
    Mat1 ret;
    memset(ret.mat,0,sizeof(ret));
    for(int i = 0; i<2; ++i)
        for(int j = 0; j<2; ++j)
            for(int k = 0; k<2; ++k)
                ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
    return ret;
}

Mat2 operator * (Mat1 A,Mat2 B)
{
    Mat2 ret;
    memset(ret.mat,0,sizeof(ret));
    for(int i = 0; i<2; ++i)
        for(int j = 0; j<1; ++j)
            for(int k = 0; k<2; ++k)
                ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
    return ret;
}

void matquickpowmod()
{
    Mat1 x;
    x.mat[0][0] = 2*a;
    x.mat[0][1] = b-a*a;
    x.mat[1][0] = 1;
    x.mat[1][1] = 0;
    Mat2 y;
    y.mat[0][0] = 2*a;
    y.mat[1][0] = 2;
    while(n){
        if(n&1)
            y = x*y;
        x = x*x;
        n >>= 1;
    }
    printf("%I64d
",(y.mat[1][0]%m+m)%m);
}

int main()
{
    while(~scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)){
        matquickpowmod();
    }
    return 0;
}