Hdu1588Gauss Fibonacci矩阵

题意：求 g(i)=k*i+b;  f(g(i)) for 0<=i<n。

 设 矩阵 A   ，  S = A^0 + A^1 + A^2 + ... + A^n 。

将这个序列分成两半 ，A^0 + A^1 + A^2 + ... + A^(n/2)      +       A^(n/2 + 1) *(A^0 + ... + A^(n/2))  + A^n  or A^0 + A^1 + A^2 + ... + A^(n/2)      +       A^(n/2 + 1) *(A^0 + ... + A^(n/2))  

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <string.h>
typedef long long LL;
LL M;
using namespace std;
struct Matrix
{
    LL m[4][4];
};

Matrix Mul(Matrix a, Matrix b)
{
    Matrix ans;
    for (LL i = 0; i < 2; i++)
    for (LL j = 0; j < 2; j++){
        ans.m[i][j] = 0;
        for (LL k = 0; k < 2; k++)
            ans.m[i][j] += a.m[i][k] * b.m[k][j];
        ans.m[i][j] %= M;
    }
    return ans;
}

Matrix add(Matrix a, Matrix b)
{
    for (LL i = 0; i < 2; i++)
    for (LL j = 0; j < 2; j++)
        a.m[i][j] += b.m[i][j], a.m[i][j] %= M;
    return a;
}

Matrix quick(Matrix a, LL b)
{
    Matrix ans;
    for (LL i = 0; i < 2; i++)
    for (LL j = 0; j < 2; j++)
        ans.m[i][j] = (i == j);
    while (b){
        if (b & 1) ans = Mul(ans, a);
        a = Mul(a, a);
        b >>= 1;
    }
    return ans;
}

Matrix solve(Matrix a, LL len)
{
    if (len == 1){
        Matrix ans;
        for (int i = 0; i<2; i++)
        for (int j = 0; j<2; j++)
            ans.m[i][j] = (i == j);
        return ans;
    }
    Matrix ans = solve(a, len >> 1);
    Matrix t = quick(a, (len >> 1));
    t = Mul(t, ans);
    ans = add(ans, t);
    if (len & 1) return add(ans, quick(a, len - 1));
    return ans;
}

void gao(LL n, LL k, LL b)
{
    Matrix ans;
    ans.m[0][0] = 1; ans.m[0][1] = 1;
    ans.m[1][0] = 1; ans.m[1][1] = 0;
    Matrix  t = quick(ans, k);
    Matrix gg = solve(t, n);
    gg = Mul(gg, quick(ans, b));
    cout << gg.m[0][1] % M << endl;
}

int main()
{
    LL n, k, b;
    while (cin >> k >> b >> n >> M){
        gao(n, k, b);
    }
    return 0;
}