ZOJ Problem Set - 3772 Calculate the Function 矩阵+线段树

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5235

因为有F(x)*A(x+2) + F(x+1) = F(x+2), 所以构造矩阵

{0         1}    {F(x)}       {F(x+1)}

{A(x+2) 0}*{F(x+1)}={F(x+2)}

用线段树维护乘号左边的矩阵的乘积，注意矩阵相乘的先后顺序，矩阵不符合交换率但符合结合律。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define ls(x) (x<<1)
#define rs(x) ((x<<1)|1)

using namespace std;

typedef long long ll;
const int maxn = 100000+10;
const int mod = 1000000007;
int a[maxn];
struct Mat
{
    int m[2][2];
    void setE()
    {
        m[0][0] = m[1][1] = 1;
        m[0][1] = m[1][0] = 0;
    }
    void setZ()
    {
        m[0][0] = m[0][1] = m[1][0] = m[1][1] = 0;
    }
    void display()
    {
        for(int i = 0; i < 2; i++){
            for(int j = 0; j < 2; j++)
            {
                printf("%d ", m[i][j]);
            }
            putchar('\n');
        }
    }
};
Mat mat[maxn * 8];

Mat operator * (const Mat &a, const Mat &b)
{
    Mat ret;
    ret.setZ();
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            for(int k = 0; k < 2; k++)
                ret.m[i][j] = ((ll)a.m[i][k]*b.m[k][j] + ret.m[i][j])%mod;
    return ret;
}

void Build(int cur, int  l, int r)
{
    int m;
    if(l == r)
    {
        mat[cur].m[0][0] = 0;
        mat[cur].m[0][1] = 1;
        mat[cur].m[1][0] = a[l];
        mat[cur].m[1][1] = 1;
    }else
    {
        m = l + ((r-l)>>1);
        Build(ls(cur), l, m);
        Build(rs(cur), m+1, r);
        mat[cur] = mat[rs(cur)] * mat[ls(cur)];
    }
}

Mat Query(int cur, int l, int r, int ql, int qr)
{
    int m;
    Mat ret;
    if(ql <= l && r <= qr)
        return mat[cur];
    else
    {
        m = l + ((r-l)>>1);
        ret.setE();
        if(ql <= m)
            ret = ret * Query(ls(cur), l, m, ql, qr);
        if(m < qr)
            ret = Query(rs(cur), m+1, r, ql, qr) * ret;
    }
    return ret;
}


int main()
{
//    freopen("data.in", "r", stdin);
    int tcas, n, m, l, r;
    Mat L, R, ret;
    scanf("%d", &tcas);
    while(tcas--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        Build(1, 1, n);
        while(m--)
        {
            scanf("%d%d", &l, &r);
            R.m[0][0] = a[l];
            R.m[0][1] = 0;
            R.m[1][0] = a[l+1];
            R.m[1][1] = 0;
            if(r - l >= 2)
            {
                L = Query(1, 1, n, l+2, r);
                ret = L * R;
                printf("%d\n", ret.m[1][0]);
            }
            else
            {
                    printf("%d\n", a[r]);
            }
        }
    }
    return 0;
}