HDU——4565So Easy!（矩阵快速幂）

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4316    Accepted Submission(s): 1402

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

 

---

难点在于这题是考数学的，只能根据它a与b的范围推出

其中Cn=ceil(a+sqrt(b))
C0=2，C1=2*a，还有一个坑点就是最后输出的答案要向正方向取模，WA好几次
代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
LL mod;
struct mat
{
    LL pos[2][2];
    mat(){memset(pos,0,sizeof(pos));}
};
inline mat operator*(const mat &a,const mat &b)
{
    mat c;
    for (int i=0; i<2; i++)
        for (int j=0; j<2; j++)
            for (int k=0; k<2; k++)
                c.pos[i][j]+=((a.pos[i][k]%mod)*(b.pos[k][j]%mod)+mod)%mod;
    return c;
}
inline mat matpow(mat a,int b)
{
    mat bas,r;
    r.pos[0][0]=r.pos[1][1]=1;
    bas=a;
    while (b!=0)
    {
        if(b&1)
            r=r*bas;
        bas=bas*bas;
        b>>=1;
    }
    return r;
}
int main(void)
{
    LL a,b,n;   
    while (~scanf("%lld%lld%lld%lld",&a,&b,&n,&mod))
    {
        mat one,t;
        one.pos[0][0]=2*a;one.pos[1][0]=2;

        t.pos[0][0]=2*a;t.pos[0][1]=-(a*a-b);
        t.pos[1][0]=1;t.pos[1][1]=0;

        t=matpow(t,n);
        one=t*one;
        printf("%lld
",(one.pos[1][0]%mod+mod)%mod);
    }
    return 0;
}