HAUToj 1284: SP教数学 (线段树+矩阵快速幂

1284: SP教数学

Description

Input

Output

对于每组数据的2操作，输出一行对1e9 + 7取模的答案

Sample Input

7 42 2 1 1 3 3 22 1 52 6 71 3 4 32 6 6

Sample Output

632

题解：

//比赛全是原题 数据要不卡的难受 要不水的暴力能把2^50的复杂度跑过去 也是服气
不想费事安安心心拉VJ不好么TAT
线段树+矩阵快速幂

出题人的解释

我们知道对于斐波那契数可以通过乘以一个矩阵来进行递推。那么对于 +x 这一区间操作，我们可以对这一区间均乘以这个矩阵：

0 11 1 

​ 所以我们可以通过维护一棵线段树来完成这些，线段树上的每一个节点存储一个 1 x 2 的矩阵 [f(n-1),f(n)] 即可。为了快速计算某个 f(n) 的值，我们可以使用矩阵快速幂。
时间复杂度： O(mlogn + (n+m)logx)

下面有两个AC代码供参考

弱者的代码

AC代码

#include <bits/stdc++.h>using namespace std;#define LL long longconst int N = 100100;const int mod = 1e9 + 7;int n, m, c[N];LL M[2][2] = {{0ll, 1ll}, {1ll, 1ll}}, V[2][2];struct node{    LL mat[2][2];    LL lz[2][2];    bool pu;}t[N << 2];#define lc x << 1#define rc x << 1 | 1void rset(LL a0[2][2]) { a0[0][0] = 1, a0[1][1] = 1, a0[1][0] = 0, a0[0][1] = 0;}void mat_mul(LL a0[2][2], LL b0[2][2], LL ret[2][2]){    LL tmp[2][2];    memset(tmp, 0, sizeof(tmp));    for (int i = 0; i <= 1; i ++)        for (int j = 0; j <= 1; j ++)            for (int k = 0; k <= 1; k ++) tmp[i][j] = (tmp[i][j] + a0[i][k] * b0[k][j] % mod) % mod;    for (int i = 0; i <= 1; i ++)        for (int j = 0; j <= 1; j ++) ret[i][j] = tmp[i][j];}void mat_pow(LL a0[2][2], int k, LL ans[2][2]){    LL tmp[2][2];    for (int i = 0; i <= 1; i ++)        for (int j = 0; j <= 1; j ++) tmp[i][j] = a0[i][j];    while(k){        if (k & 1) mat_mul(ans, tmp, ans);        mat_mul(tmp, tmp, tmp);        k >>= 1;    }}void up(int x){    t[x].mat[0][0] = (t[lc].mat[0][0] + t[rc].mat[0][0]) % mod;    t[x].mat[1][0] = (t[lc].mat[1][0] + t[rc].mat[1][0]) % mod;}void pushdown(int x){    if (!t[x].pu) return ;    t[lc].pu = true, t[rc].pu = true, t[x].pu = false;    mat_mul(t[x].lz, t[lc].mat, t[lc].mat);    mat_mul(t[x].lz, t[rc].mat, t[rc].mat);    mat_mul(t[x].lz, t[lc].lz, t[lc].lz);    mat_mul(t[x].lz, t[rc].lz, t[rc].lz);    rset(t[x].lz);}void build(int x, int l, int r){    rset(t[x].lz); rset(t[x].mat); t[x].pu = false;    if (l == r){        mat_pow(M, c[l] - 1, t[x].mat);        t[x].mat[0][0] += t[x].mat[0][1], t[x].mat[0][0] %= mod;        t[x].mat[1][0] += t[x].mat[1][1], t[x].mat[1][0] %= mod;        return;    }    int mid = (l + r) >> 1;    build(lc, l, mid); build(rc, mid + 1, r); up(x);}void modify(int x, int l, int r, int a, int b){    if (a <= l && r <= b){        t[x].pu = true;        mat_mul(t[x].lz, V, t[x].lz);        mat_mul(V, t[x].mat, t[x].mat);        return ;    }    pushdown(x);    int mid = (l + r) >> 1;    if (a <= mid) modify(lc, l, mid, a, b);    if (b > mid) modify(rc, mid + 1, r, a, b);    up(x);}LL query(int x, int l, int r, int a, int b){    if (a <= l && r <= b) return t[x].mat[0][0];    pushdown(x);    int mid = (l + r) >> 1;    LL ret = 0;    if (a <= mid) ret = (ret + query(lc, l, mid, a, b)) % mod;    if (b > mid) ret = (ret + query(rc, mid + 1, r, a, b)) % mod;    return ret;}int read(){    int x = 0; char ch = getchar();    while (ch < '0' || ch > '9') ch = getchar();    while (ch >= '0' && ch <= '9'){ x = x * 10 + ch - '0'; ch = getchar(); }    return x;}int main(){    n = read(), m = read();    for (int i = 1; i <= n; i ++) c[i] = read();    build(1, 1, n);    int ty, l, r, v;    for (int i = 1; i <= m; i ++){        ty = read();        if (ty == 1) {            l = read(), r = read(), v = read();            rset(V); mat_pow(M, v, V);            modify(1, 1, n, l, r);        }        else {            l = read(), r = read();            printf("%d\n", query(1, 1, n, l, r));        }    }    return 0;}

出题人的标程

#include <bits/stdc++.h>#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1using namespace std;typedef long long ll;const ll MOD = 1e9 + 7;const int MAXN = 1e5 + 10;struct node {    ll x, y;}sum[MAXN << 2];struct mat{    ll p[3][3], sz;    mat operator + (const mat &x) {        mat tp; tp.sz = sz;        for (int i = 1; i <= sz; i++)            for (int j = 1; j <= sz; j++)                tp.p[i][j] = (p[i][j] + x.p[i][j]) % MOD;        return tp;    }    mat operator * (const mat &x) {        mat tp; tp.sz = sz;        for (int i = 1; i <= sz; i++)            for (int j = 1; j <= sz; j++) {                tp.p[i][j] = 0;                for (int k = 1; k <= sz; k++)                    tp.p[i][j] = (tp.p[i][j] + p[i][k] * x.p[k][j] % MOD) % MOD;            }        return tp;    }}unit, Am;ll a[MAXN];mat lazy[MAXN << 2];mat pow_mod(mat a, ll n) {    mat res = unit;    while (n) {        if (n & 1) res = res * a;        a = a * a;        n >>= 1;    }    return res;}void init() {    memset(&unit, 0, sizeof(unit));    memset(&Am, 0, sizeof(Am));    unit.sz = Am.sz = 2;    for (int i = 1; i <= 2; i++) unit.p[i][i] = 1;    Am.p[1][1] = Am.p[1][2] = Am.p[2][1] = 1;}bool Is_unit(mat x) {    if (x.p[1][1] || x.p[2][2]) return false;    return true;}void push_up(int rt) {    sum[rt].x = (sum[rt << 1].x + sum[rt << 1 | 1].x) % MOD;    sum[rt].y = (sum[rt << 1].y + sum[rt << 1 | 1].y) % MOD;}void push_down(int rt) {    ll x, y;    if (!Is_unit(lazy[rt])) {        lazy[rt << 1] = lazy[rt << 1] * lazy[rt];        lazy[rt << 1 | 1] = lazy[rt << 1 | 1] * lazy[rt];        x = (sum[rt << 1].x * lazy[rt].p[1][1] % MOD + sum[rt << 1].y * lazy[rt].p[1][2] % MOD) % MOD;        y = (sum[rt << 1].x * lazy[rt].p[2][1] % MOD + sum[rt << 1].y * lazy[rt].p[2][2] % MOD) % MOD;        sum[rt << 1] = (node){x, y};        x = (sum[rt << 1 | 1].x * lazy[rt].p[1][1] % MOD + sum[rt << 1 | 1].y * lazy[rt].p[1][2] % MOD) % MOD;        y = (sum[rt << 1 | 1].x * lazy[rt].p[2][1] % MOD + sum[rt << 1 | 1].y * lazy[rt].p[2][2] % MOD) % MOD;        sum[rt << 1 | 1] = (node){x, y};        lazy[rt] = unit;    }}void build(int l, int r, int rt) {    lazy[rt] = unit;    if (l == r) {        if (a[l] == 1) sum[rt] = (node) {1, 0};        else if (a[l] == 2) sum[rt] = (node) {1, 1};        else {            mat tmp = pow_mod(Am, a[l] - 2);            sum[rt].x = (1 * tmp.p[1][1] + 1 * tmp.p[1][2]) % MOD;            sum[rt].y = (1 * tmp.p[2][1] + 1 * tmp.p[2][2]) % MOD;        }        return;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    push_up(rt);}ll query(int L, int R, int l, int r, int rt) {    if (L <= l && r <= R) return sum[rt].x;    push_down(rt);    int m = (l + r) >> 1;    ll res = 0;    if (L <= m) res = (res + query(L, R, lson)) % MOD;    if (R > m) res = (res + query(L, R, rson)) % MOD;    return res;}void update(int L, int R, mat add, int l, int r, int rt) {    if (L <= l && r <= R) {        lazy[rt] = lazy[rt] * add;        ll x = (sum[rt].x * add.p[1][1] % MOD + sum[rt].y * add.p[1][2] % MOD) % MOD;        ll y = (sum[rt].x * add.p[2][1] % MOD + sum[rt].y * add.p[2][2] % MOD) % MOD;        sum[rt] = (node){x, y};        return;    }    push_down(rt);    int m = (l + r) >> 1;    if (L <= m) update(L, R, add, lson);    if (R > m) update(L, R, add, rson);    push_up(rt);}int main() {    //freopen("in0.in", "r", stdin);    init();    int n, m;    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; i++)        scanf("%lld", &a[i]);    build(1, n, 1);    while (m--) {        int op, l, r;        ll x;        scanf("%d", &op);        if (op == 1) {            scanf("%d%d%lld", &l, &r, &x);            mat add = pow_mod(Am, x);            update(l, r, add, 1, n, 1);        }        else {            scanf("%d%d", &l, &r);            printf("%lld\n", query(l, r, 1, n, 1));        }    }    return 0;}