ZOJ 3772 矩阵 + 线段树

最新推荐文章于 2019-04-30 14:56:13 发布

原创最新推荐文章于 2019-04-30 14:56:13 发布 · 526 阅读

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#zoj #线段树 #矩阵

数据结构-线段树（Segment tree）同时被 2 个专栏收录

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数论-矩阵快速幂&&快速幂

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本文介绍了解决ZOJ3772问题的方法，采用节点存储矩阵的方式，并考虑了方向因素。通过ZKW线段树实现矩阵的构建与查询，提供了完整的AC代码。

传送门：zoj 3772

题解

还是节点存储矩阵，不过这题各哥节点基矩阵不同，所以要考虑方向，wa + 3, pe + 1

AC code:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

#define mid ((l + r) >> 1)
#define ls rt << 1, l, mid
#define rs rt << 1 | 1, mid + 1, r
typedef long long LL;

const LL MOD = 1000000007;
const LL maxn = 100000 + 5;

int n, M;

struct Matrix{

    LL v[2][2];

    void clr() {
        memset(v, 0, sizeof(v));
    }

    void E() {
        clr();
        for (int i = 0; i < 2; ++i) v[i][i] = 1;
    }

    void init(LL a, LL b, LL c, LL d) {
        v[0][0] = a;
        v[0][1] = b;
        v[1][0] = c;
        v[1][1] = d;
    }

    void debug() {
        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 2; ++j)
                cout << v[i][j] << (j == 1 ? "\n" : " ");
    }

    Matrix operator * (Matrix b) {
        Matrix res;
        res.clr();

        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 2; ++j)
                for(int k = 0; k < 2; ++k)
                    res.v[i][j] = (res.v[i][j] + v[i][k] * b.v[k][j]) % MOD;

        return res;
    }

    //Matrix 
}a[maxn << 2];

void build() {//zkw线段树
    LL x;

    for (M = 1; M <= n + 1; M <<= 1);
    for (int i = M + 1; i <= M + n; ++i) {
        scanf("%I64d", &x);
        a[i].init(0, x, 1, 1);
    }

    for (int i = M - 1; i; --i) {
        a[i] = a[i << 1] * a[i << 1 | 1];
    }
}

LL query(int l, int r) {
    int id = l + M;
    l += M + 1;//查询[l + 2, r]的矩阵积， l += M - 1 + 2;
    r += M + 1;

    Matrix tmp, la, ra;
    la.E();
    ra.E();
    //tmp.E();

    while (r ^ l ^ 1) {
        if (~l & 1) la = la * a[l + 1];//方向问题， 后来才反应过来
        if (r & 1) ra = a[r - 1] * ra;
        l >>= 1;
        r >>= 1;
    }
    tmp = la * ra;
    LL res = (a[id].v[0][1] * tmp.v[0][1] + a[id + 1].v[0][1] * tmp.v[1][1]) % MOD;
    if (res < 0) res += MOD;

    return res;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif 

    cin.tie(0);
    cin.sync_with_stdio(false);

    int T, q, l, r;
    scanf("%d", &T);

    while (T--) {
        for (int i = 1; i < maxn << 2; ++i) a[i].E();
        scanf("%d%d", &n, &q);
        build();

        while (q--) {
            scanf("%d%d", &l, &r);
            if(r - l >= 2)  printf("%d\n", query(l, r));
            else printf("%d\n", a[r + M].v[0][1]);
        }
    }

    return 0;
}