【HDU 1588】Gauss Fibonacci（矩阵快速幂+二分）

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3144 Accepted Submission(s): 1319

Problem Description

Without expecting, Angel replied quickly.She says: “I’v heard that you’r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. ”

How good an opportunity that Gardon can not give up! The “Problem GF” told by Angel is actually “Gauss Fibonacci”.

As we know ,Gauss is the famous mathematician who worked out the sum from 1 to 100 very quickly, and Fibonacci is the crazy man who invented some numbers.

Arithmetic progression:

g(i)=k*i+b;

We assume k and b are both non-nagetive integers.

Fibonacci Numbers:

f(0)=0 ，f(1)=1 ，f(n)=f(n-1)+f(n-2) (n>=2)

The Gauss Fibonacci problem is described as follows:

Given k,b,n ,calculate the sum of every f(g(i)) for 0<=i<n

The answer may be very large, so you should divide this answer by M and just output the remainder instead.

Input

The input contains serveral lines. For each line there are four non-nagetive integers: k,b,n,M

Each of them will not exceed 1,000,000,000.

Output

For each line input, out the value described above.

Sample Input

2 1 4 100

2 0 4 100

Sample Output

21

12

[题意][求∑n−1i=0 f[k*i+b] % p 的值（f代表斐波那契数列）]

【题解】【矩阵快速幂】

A={{1,1},{1,0}},B={{1,0},{0,1}}

∑n−1i=0 f[k*i+b]

=Ab+Ak+b+A2k+b+…+A(n−1)∗k+b

=Ab∗(1+Ak+A2∗k+A3∗k+…+A(n−1)∗k)

设：B=Ak

∴ =Ab∗(B0+B1+B2+B3+…+Bn−1)

由本题的范围可以得知，不可能直接枚举，所以需要优化时间，二分+递归无疑是最显然的方法……

又∵ 对于矩阵S（m）=F1+F2+F3+……+Fm

当m为偶数时，S(m)=(1+Fm/2)∗(F+F2+F3+…+Fm/2)

当m为奇数时，S(m)=F+(F+Fm/2+1)∗(F+F2+…+Fm/2)

∴ 在本题中，式子最终化为 ∑n−1i=0 f[k*i+b]=Ab∗(B0+S)

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
struct node{
ll k[2][2];
}a,e,t;
int n,k,b;
ll mod;

inline node jc(node a1,node b1)
{
int i,j,l;
node c;
c.k[0][0]=c.k[0][1]=c.k[1][0]=c.k[1][1]=0;
for(i=0;i<=1;++i)
for(j=0;j<=1;++j)
for(l=0;l<=1;++l)
c.k[i][j]=(c.k[i][j]+a1.k[i][l]*b1.k[l][j]%mod)%mod;
return c;
}
inline node add(node a1,node b1)
{
int i,j;
node c;
for(i=0;i<=1;++i)
for(j=0;j<=1;++j)
c.k[i][j]=(a1.k[i][j]+b1.k[i][j])%mod;
return c;
}
node poww(node x,int p)
{
node sum;
sum.k[0][0]=sum.k[1][1]=1; sum.k[0][1]=sum.k[1][0]=0;
node c=x;
while(p)
{
if (p&1) sum=jc(sum,c);
p>>=1;
c=jc(c,c);
}
return sum;
}
node check(int h)
{
if(h==1) return t;
node sum=poww(t,((h+1)/2));
node s1=check(h/2);
if(h%2) return add(t,jc(add(t,sum),s1));
else return jc(add(e,sum),s1);
}

int main()
{
a.k[0][0]=a.k[0][1]=a.k[1][0]=1; a.k[1][1]=0;
e.k[0][0]=e.k[1][1]=1; e.k[0][1]=e.k[1][0]=0;
while((scanf("%d%d%d%lld",&k,&b,&n,&mod))==4)
{
node t1=poww(a,b);
t=poww(a,k);
node ans=jc(t1,add(e,check(n-1)));
printf("%lld\n",ans.k[0][1]);
}
}