HDU6395 Sequence(矩阵快速幂+数论分块)

题意：

F(1)=A，F(2)=B，F(n)=C*F(n-2)+D*F(n-1)+P/n

给定ABCDPn，求F(n) mod 1e9+7

思路：

P/n在一段n里是不变的，可以数论分块，再在每一段里用矩阵快速幂

debug了一下午。。

坑点：

1.数论分块的写法要注意，已更新

2.矩阵乘法在赋值回去的时候记得模一下

3.矩阵相乘不可逆，注意看一下

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
void prt(ll b[4][4]){
    for(int i = 1; i <= 3; i++){
        for(int j = 1; j <= 3; j++){
            printf("%3I64d ", b[i][j]);
        }
        printf("
");
    }
    return;
}
void mtpl(ll a[4][4], ll b[4][4], ll s[4][4]){
    ll tmp[4][4];
    mem(tmp, 0);
    for(int i = 1; i <= 3; i++){
        for(int j = 1; j <= 3; j++){
            
            for(int k = 1; k <= 3; k++){
                tmp[i][j] += (a[i][k]*b[k][j])%mod;
            }
        }
    }
    for(int i = 1; i <= 3; i++){
        for(int j = 1; j <= 3; j++){
            s[i][j] = tmp[i][j]%mod;
        }
    }
    return;
}ll a, b, c, d, p, n;
ll B[4][4];
void fp(ll n, ll tmp){
    ll A[4][4];
    mem(A, 0);
    A[1][1] = d;
    A[1][2] = c;
    A[2][1] = 1;
    A[1][3] = tmp;
    A[3][3] = 1;
    //prt(A);
    while(n){
        if(n&1) mtpl(A,B,B);
        mtpl(A,A,A);
        n>>=1;
    }
    return;
}
int main() {
    int T;
    scanf("%d",&T);
    while(T--){
        
        scanf("%I64d %I64d %I64d %I64d %I64d %I64d", &a, &b, &c, &d, &p, &n);
        
        if(n==1){
            printf("%I64d
", a);
            continue;
        }
        if(n==2){
            printf("%I64d
", b);
            continue;
        }
        
        mem(B, 0);
        for(int i = 1; i <= 3; i++)B[i][i]=1;
        B[1][1] = b;
        B[2][1] = a;
        B[3][1] = 1;
        ll l , r;
        for(l = 3, r = 0; l <= n; l = r + 1){
            if(p/l) r = min(p, p/(p/l));
            else r = n;
            ll tmp = p/l;
            fp(r-l+1,tmp);
        }
        

        printf("%I64d
", B[1][1]);
    }
    return 0;
}
/*
4
1 2 1 2 5 3
1 2 1 2 3 5
1 2 3 4 7 4
2 3 3 3 3 4

 */