So Easy!（矩阵快速幂）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=4565

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3370    Accepted Submission(s): 1080


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

 

Source

2013
ACM-ICPC长沙赛区全国邀请赛——题目重现

 

AC code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#define LL long long
#define MAXN 1000010
using namespace std;
const int INF=0x3f3f3f3f;
//----以下为矩阵快速幂模板-----//
int mod=1024;
const int NUM=11;//定义矩阵能表示的最大维数
int N;//N表示矩阵的维数，以下的矩阵加法、乘法、快速幂都是按N维矩阵运算的
struct Mat{//矩阵的类
LL a[NUM][NUM];
void init()//将其初始化为单位矩阵
{
memset(a,0,sizeof(a));
for(int i=0;i<NUM;i++)
{
a[i][i]=1;
}
}
};
Mat add(Mat a,Mat b)//(a+b)%mod 矩阵加法
{
Mat ans;
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
ans.a[i][j]=a.a[i][j]+b.a[i][j];
ans.a[i][j]%=mod;
}
}
return ans;
}
Mat mul(Mat a,Mat b) //(a*b)%mod 矩阵乘法
{
Mat ans;
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
ans.a[i][j]=0;
for(int k=0;k<N;k++)
{
ans.a[i][j]+=a.a[i][k]*b.a[k][j];
}
ans.a[i][j]%=mod;
}
}
return ans;
}
Mat power(Mat a,int num)//(a^n)%mod 矩阵快速幂
{
Mat ans;
ans.init();
while(num)
{
if(num&1)
{
ans=mul(ans,a);
}
num>>=1;
a=mul(a,a);
}
return ans;
}
Mat pow_sum(Mat a,int num)//(a+a^2+a^3....+a^n)%mod 矩阵的幂和
{
int m;
Mat ans,pre;
if(num==1)
return a;
m=num/2;
pre=pow_sum(a,m);
ans=add(pre,mul(pre,power(a,m)));
if(num&1)
ans=add(ans,power(a,num));
return ans;
}
void output(Mat a)//输出矩阵
{
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
printf("%d%c",a.a[i][j],j==N-1?'\n':' ');
}
}
}
//----以上为矩阵快速幂模板-----//
int main()
{
int t,k,ans,i,j,A,B,n,m;
int T;
while(scanf("%d%d%d%d",&A,&B,&n,&m)!=EOF)
{
if(n==1){
cout << (2*A)%m << endl;
continue;
}
Mat a,f;
N=2;
a.a[0][0]=A;
a.a[0][1]=B;
a.a[1][0]=1;
a.a[1][1]=A;
/*k=n%m;//k为循环节
if(k==0)
k=m;*/
mod=m;
f=power(a,n-1);
int an=f.a[0][0]*A+f.a[0][1]*1;
printf("%d\n",(2*an)%mod);
}
return 0;
}