HDU 1588 Gauss Fibonacci(斐波那契数列-矩阵)
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1588
题意:函数g和f定义如下。给出k,b,n,M，求s(n)=sigama(f(g(i)))0<=i<n。
（1）g(i)=k*i+b;
（2）f(0)=0，f(1)=1，f(n)=f(n-1)+f(n-2) (n>=2)
思路：构造矩阵a={0,1,1,1}，则a^n=f(n)。s(n)=a^b*(a^0+a^k+a^(2k)+……+a^((n-1)*k))。
#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <string>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define abs(x) ((x)>=0?(x):-(x))
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define clr(x,y) memset(x,y,sizeof(x))
#define CLR(x) x.clear()
#define ph(x) push(x)
#define pb(x) push_back(x)
#define Len(x) x.length()
#define SZ(x) x.size()
#define PI acos(-1.0)
#define sqr(x) ((x)*(x))
#define FOR0(i,x) for(i=0;i<x;i++)
#define FOR1(i,x) for(i=1;i<=x;i++)
#define FOR(i,a,b) for(i=a;i<=b;i++)
#define DOW0(i,x) for(i=x;i>=0;i--)
#define DOW1(i,x) for(i=x;i>=1;i--)
#define DOW(i,a,b) for(i=a;i>=b;i--)
using namespace std;
void RD(int &x){scanf("%d",&x);}
void RD(i64 &x){scanf("%lld",&x);}
void RD(u64 &x){scanf("%llu",&x);}
void RD(u32 &x){scanf("%u",&x);}
void RD(double &x){scanf("%lf",&x);}
void RD(int &x,int &y){scanf("%d%d",&x,&y);}
void RD(i64 &x,i64 &y){scanf("%lld%lld",&x,&y);}
void RD(u64 &x,u64 &y){scanf("%llu%llu",&x,&y);}
void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}
void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}
void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}
void RD(i64 &x,i64 &y,i64 &z){scanf("%lld%lld%lld",&x,&y,&z);}
void RD(u64 &x,u64 &y,u64 &z){scanf("%llu%llu%llu",&x,&y,&z);}
void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}
void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}
void RD(char &x){x=getchar();}
void RD(char *s){scanf("%s",s);}
void RD(string &s){cin>>s;}
void PR(int x) {printf("%d\n",x);}
void PR(i64 x) {printf("%I64d\n",x);}
void PR(u64 x) {printf("%llu\n",x);}
void PR(u32 x) {printf("%u\n",x);}
void PR(double x) {printf("%lf\n",x);}
void PR(char x) {printf("%c\n",x);}
void PR(char *x) {printf("%s\n",x);}
void PR(string x) {cout<<x<<endl;}
int mod;
const int INF=2000000000;
const int N=4;
class Matrix
{
public:
i64 a[N][N];
int n;
void init(int x)
{
int i;
if(!x) clr(a,0);
else
{
clr(a,0);
FOR0(i,2) a[i][i]=1;
}
}
void read(int n)
{
int i,j;
FOR1(i,n) FOR1(j,n)
{
RD(a[i][j]);
a[i][j]%=mod;
}
}
Matrix operator+(Matrix b)
{
int i,j;
Matrix c;
FOR0(i,2) FOR0(j,2) c.a[i][j]=(a[i][j]+b.a[i][j])%mod;
return c;
}
Matrix operator+(int x)
{
int i;
Matrix c=*this;
FOR0(i,2) (c.a[i][i]+=x)%=mod;
return c;
}
Matrix operator*(Matrix b)
{
int i,j,k;
Matrix p;
p.init(0);
FOR0(i,2) FOR0(j,2) FOR0(k,2)
{
(p.a[i][j]+=(i64)a[i][k]*b.a[k][j]%mod)%=mod;
}
return p;
}
Matrix power(int t)
{
Matrix ans,p=*this;
ans.init(1);
while(t)
{
if(t&1) ans=ans*p;
p=p*p;
t>>=1;
}
ans.n=n;
return ans;
}
void print()
{
int i,j;
FOR0(i,2)
{
FOR0(j,2) printf("%d ",a[i][j]);
puts("");
}
}
};
Matrix a;
int k,b,n,M;
Matrix cal(Matrix a,int n,int k)
{
Matrix ans;
ans.init(1);
if(n==1) return ans;
if(n==2) return ans+a.power(k);
if(n&1)
{
ans=(a.power(n/2).power(k)+1)*cal(a,n/2,k)+a.power(n-1).power(k);
}
else
{
ans=(a.power(n/2).power(k)+1)*cal(a,n/2,k);
}
return ans;
}
int main()
{
while(scanf("%d%d%d%d",&k,&b,&n,&M)!=-1)
{
a.a[0][0]=0;
a.a[0][1]=a.a[1][0]=a.a[1][1]=1;
mod=M;
Matrix ans=cal(a,n,k)*a.power(b);
PR(ans.a[0][1]);
}
return 0;
}