POJ——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): 3642 Accepted Submission(s): 1185

Problem Description
　　A sequence Sn 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 Sn.
　　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 Sn.

Sample Input
2 3 1 2013
2 3 2 2013
2 2 1 2013

Sample Output
4
14
4

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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 0x3f3f3f3fLL 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\n",(one.pos[1][0]%mod+mod)%mod);    }    return 0;}