线段树+矩阵快速幂 Codeforces Round #373 (Div. 2) E

http://codeforces.com/contest/719/problem/E

题目大意：给你一串数组a，a[i]表示第i个斐波那契数列，有如下操作

①对[l,r]区间+一个val ②求出[l,r]区间的和。

定义区间的和为该区间内每个a[i]所对应的斐波那契数列的和。

思路：线段树保存区间val，和区间更新，用矩阵快速幂求解复杂度是m*logn*logk

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha
")
const int maxn = 1e5 + 5;
const LL mod = 1e9 + 7;
struct Node{
    LL mat[2][2];
    void reset(){memset(mat, 0, sizeof(mat));}
    void getone(){
        reset();
        mat[0][0] = mat[1][1] = 1;
    }
};

inline Node mul(Node A, Node B){
    Node ans; ans.reset();
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
            for (int k = 0; k < 2; k++)
                ans.mat[i][j] = (ans.mat[i][j] + A.mat[i][k] * B.mat[k][j]) % mod;
    return ans;
}

inline Node add(Node a, Node b){
    Node ans; ans.reset();
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
            ans.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % mod;
    return ans;
}

inline Node kpow(LL k){
    Node ans; ans.reset();
    ans.mat[0][0] = ans.mat[1][1] = 1;
    Node A;
    A.mat[0][0] = A.mat[1][0] = A.mat[0][1] = 1;
    A.mat[1][1] = 0;
    while (k){
        if (k & 1) ans = mul(ans, A);
        A = mul(A, A);
        k >>= 1;
    }
    return ans;
}

struct Mytree{
    Node sum;
    Node lazyk;
}tree[maxn << 2];

int n, m;

inline void push_up(int o){
    int lb = o << 1, rb = o << 1 | 1;
    tree[o].sum = add(tree[lb].sum, tree[rb].sum);
}

void build_tree(int l, int r, int o){
    if (l == r){
        LL k; scanf("%lld", &k);
        tree[o].sum = kpow(k);
        tree[o].lazyk.getone();
        return ;
    }
    tree[o].sum.reset(); tree[o].lazyk.getone();
    int mid = (l + r) / 2;
    if (l <= mid) build_tree(l, mid, o << 1);
    if (r > mid) build_tree(mid + 1, r, o << 1 | 1);
    push_up(o);
}

inline void push_down(int o){
    int lb = o << 1, rb = o << 1 | 1;
    tree[lb].sum = mul(tree[lb].sum, tree[o].lazyk);
    tree[lb].lazyk = mul(tree[lb].lazyk, tree[o].lazyk);

    tree[rb].sum = mul(tree[rb].sum, tree[o].lazyk);
    tree[rb].lazyk = mul(tree[rb].lazyk, tree[o].lazyk);

    tree[o].lazyk.getone();
}

void update(int l, int r, int ql, int qr, Node k, int o){
    if (ql <= l && qr >= r){
        tree[o].sum = mul(tree[o].sum, k);
        tree[o].lazyk = mul(tree[o].lazyk, k);
        return ;
    }
    int mid = (l + r) / 2;
    push_down(o);
    if (ql <= mid) update(l, mid, ql, qr, k, o << 1);
    if (qr > mid) update(mid + 1, r, ql, qr, k, o << 1 | 1);
    push_up(o);
    return ;
}

Node query(int l, int r, int ql, int qr, int o){
    if (ql <= l && qr >= r){
        return tree[o].sum;
    }
    push_down(o);
    Node ans; ans.reset();
    int mid = (l + r) / 2;
    if (ql <= mid) ans = add(query(l, mid, ql, qr, o << 1), ans);
    if (qr > mid) ans = add(query(mid + 1, r, ql, qr, o << 1 | 1), ans);
    push_up(o);
    return ans;
}

int main(){
    scanf("%d%d", &n, &m);
    build_tree(1, n, 1);
    for (int i = 0; i < m; i++){
        int ty; scanf("%d", &ty);
        if (ty == 1){
            int l, r; LL k;
            scanf("%d%d%lld", &l, &r, &k);
            update(1, n, l, r, kpow(k), 1);
        }
        else if (ty == 2){
            int l, r; scanf("%d%d", &l, &r);
            Node ans = query(1, n, l, r, 1);
            printf("%lld
", ans.mat[1][0]);
        }
    }
    return 0;
}