Codeforces 719 E. Sasha and Array (线段树+矩阵运算)

题目链接：http://codeforces.com/contest/719/problem/E

题意：操作1将[l, r] + x;

        操作2求f[l] + ... + f[r];

题解：注意矩阵可以是a*(b + c) = a*b + a*c ，还有update的时候传入矩阵

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e5 + 5;
LL a[N], mod = 1e9 + 7;
struct data {
    LL mat[3][3];
    data() {
        mat[1][1] = mat[2][1] = mat[1][2] = 1;
        mat[2][2] = 0;
    }
    void init() {
        mat[1][1] = mat[2][2] = 1;
        mat[1][2] = mat[2][1] = 0;
    }
}op;
struct SegTree {
    int l, r;
    data lazy;
    data val;  
}T[N << 2];

data operator +(const data &a, const data &b) {
    data ans;
    for(int i = 1; i <= 2; ++i) {
        for(int j = 1; j <= 2; ++j) {
            ans.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % mod;
        }
    }
    return ans;
}

data operator *(const data &a, const data &b) {
    data ans;
    for(int i = 1; i <= 2; ++i) {
        for(int j = 1; j <= 2; ++j) {
            ans.mat[i][j] = 0;
            for(int k = 1; k <= 2; ++k) {
                ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j] % mod) % mod;
            }
        }
    }
    return ans;
}

data operator ^(data a, LL n) {
    data ans;
    for(int i = 1; i <= 2; ++i) {
        for(int j = 1; j <= 2; ++j) {
            ans.mat[i][j] = (i == j);
        }
    }
    while(n) {
        if(n & 1)
            ans = ans * a;
        a = a * a;
        n >>= 1;
    }
    return ans;
}

bool cmp(const data &a) {
    for(int i = 1; i <= 2; ++i) {
        for(int j = 1; j <= 2; ++j) {
            if(a.mat[i][j] != op.mat[i][j])
                return false;
        }
    }
    return true;
}

inline void pushdown(int p) {
    if(!cmp(T[p].lazy)) {
        T[p << 1].lazy = T[p].lazy * T[p << 1].lazy;
        T[(p << 1)|1].lazy = T[p].lazy * T[(p << 1)|1].lazy;
        T[p << 1].val = T[p << 1].val * T[p].lazy;
        T[(p << 1)|1].val = T[(p << 1)|1].val * T[p].lazy;
        T[p].lazy.init();
    }
}

void build(int p, int l, int r) {
    int mid = (l + r) >> 1;
    T[p].l = l, T[p].r = r, T[p].lazy.init();
    if(l == r) {
        scanf("%lld", a + l);
        T[p].val = T[p].val ^ (a[l] - 1);
        return ;
    }
    build(p << 1, l, mid);
    build((p << 1)|1, mid + 1, r);
    T[p].val = T[p << 1].val + T[(p << 1)|1].val;
}

void update(int p, int l, int r, data add) {
    int mid = (T[p].l + T[p].r) >> 1;
    if(T[p].l == l && T[p].r == r) {
        T[p].lazy = add * T[p].lazy;
        T[p].val = T[p].val * add;
        return ;
    }
    pushdown(p);
    if(r <= mid) {
        update(p << 1, l, r, add);
    } else if(l > mid) {
        update((p << 1)|1, l, r, add);
    } else {
        update(p << 1, l, mid, add);
        update((p << 1)|1, mid + 1, r, add);
    }
    T[p].val = T[p << 1].val + T[(p << 1)|1].val;
}

LL query(int p, int l, int r) {
    int mid = (T[p].l + T[p].r) >> 1;
    if(T[p].l == l && T[p].r == r) {
        return T[p].val.mat[1][1];
    }
    pushdown(p);
    if(r <= mid) {
        return query(p << 1, l, r);
    } else if(l > mid) {
        return query((p << 1)|1, l, r);
    } else {
        return (query(p << 1, l, mid) + query((p << 1)|1, mid + 1, r)) % mod;
    }
}

int main()
{
    op.init();
    int n, m;
    scanf("%d %d", &n, &m);
    build(1, 1, n);
    int c, l, r;
    LL add;
    while(m--) {
        scanf("%d %d %d", &c, &l, &r);
        if(c == 1) {
            scanf("%lld", &add);
            data temp;
            update(1, l, r, temp ^ add);
        } else {
            printf("%lld\n", query(1, l, r));
        }
    }
    return 0;
}