HDU 4565 矩阵快速幂

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3343    Accepted Submission(s): 1070

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
#include<cstdio>#include<iostream>using namespace std;long long c[2][2]={0},d[2]={0};int main(){long long a=0,b=0,n=0,m=0;while(scanf("%lld%lld%lld%lld",&a,&b,&n,&m)!=EOF){c[0][0]=2*a;        c[0][1]=b-a*a;        c[1][0]=1;        c[1][1]=0;        d[0]=2*a;        d[1]=2;        while(n)        {        long long x=0,y=0;        long long p=0,q=0;        if(n&1)            {                x=(c[0][0]*d[0]+c[0][1]*d[1])%m;                y=(c[1][0]*d[0]+c[1][1]*d[1])%m;                d[0]=x;                d[1]=y;            }            n=n/2;            x=(c[0][0]*c[0][0]+c[0][1]*c[1][0])%m;            y=(c[0][0]*c[0][1]+c[0][1]*c[1][1])%m;            p=(c[1][0]*c[0][0]+c[1][1]*c[1][0])%m;            q=(c[1][0]*c[0][1]+c[1][1]*c[1][1])%m;            c[0][0]=x;            c[0][1]=y;            c[1][0]=p;            c[1][1]=q;}printf("%lld\n",(d[1]+m)%m);}return 0;}