Codeforces Round #373 (Div. 2) E. Sasha and Array

题目链接

分析：矩阵快速幂+线段树 斐波那契数列的计算是矩阵快速幂的模板题，这个也没什么很多好解释的，学了矩阵快速幂应该就知道的东西= =这道题比较巧妙的在于需要用线段树来维护矩阵，达到快速查询区间斐波那契数列和的目的。这道题极为卡常数，我TLE了不知道多少发，才在赛后过了这道题。

我尝试下来，发现矩阵乘法的写法极为重要，我就是因为用了三层循环来写矩阵乘法导致了悲剧的TLE，一直卡在了第17组数据。我百度了网上别的矩阵快速幂的写法才过了这道题。
还有涨智识的地方是不要随意memset。我原来的矩阵快速幂的模板里面在构造函数的时候会memset元素为0，去掉了这句话，代码快了1s。。。

毕竟萌新还是太嫩，没有很多老司机的经验。。貌似bc87场的第三题也是卡了memset的常数。我也TLE了一发。

/*****************************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <map>#include <set>#include <ctime>#include <stack>#include <queue>#include <cmath>#include <string>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <sstream>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define   offcin        ios::sync_with_stdio(false)#define   sigma_size    26#define   lson          l,m,v<<1#define   rson          m+1,r,v<<1|1#define   slch          v<<1#define   srch          v<<1|1#define   sgetmid       int m = (l+r)>>1#define   LL            long long#define   ull           unsigned long long#define   mem(x,v)      memset(x,v,sizeof(x))#define   lowbit(x)     (x&-x)#define   bits(a)       __builtin_popcount(a)#define   mk            make_pair#define   pb            push_back#define   fi            first#define   se            secondconst int    INF    = 0x3f3f3f3f;const LL     INFF   = 1e18;const double pi     = acos(-1.0);const double inf    = 1e18;const double eps    = 1e-9;const LL     mod    = 1e9+7;const int    maxmat = 10;const ull    BASE   = 31;/*****************************************************/const int maxn = 1e5 + 5;int a[maxn];int N, M;struct Mat {    int N;    LL a[2][2];    Mat (int _n = 2) : N(_n) {        //这里我原来会memset    }    void Debug() {        for (int i = 0; i < N; i ++) {            for (int j = 0; j < N; j ++)                cout<<a[i][j]<<" ";            cout<<endl;        }    }    // Mat operator *(const Mat &rhs) const { //悲剧的三重循环写法    //     Mat c(N);    //     for (int i = 0; i < N; i ++)     //         for (int k = 0; k < N; k ++) {    //             if (!a[i][k]) continue;    //             for (int j = 0; j < N; j ++)    //                 c.a[i][j] = (c.a[i][j] + a[i][k] * rhs.a[k][j] % mod) % mod;    //         }    //     return c;    // }    Mat operator *(const Mat &rhs) const {   //同时也能能够避免三重循环需要先memset矩阵的时间复杂度        Mat ret;        ret.a[0][0] = (1ll * a[0][0] * rhs.a[0][0] + 1ll * a[0][1] * rhs.a[1][0]) % mod;        ret.a[1][0] = ret.a[0][1] = (1ll * a[0][0] * rhs.a[0][1] + 1ll * a[0][1] * rhs.a[1][1]) % mod;        ret.a[1][1] = (1ll * a[1][0] * rhs.a[0][1] + 1ll * a[1][1] * rhs.a[1][1]) % mod;        return ret;    }    Mat operator +(const Mat &rhs) const {        Mat c(N);        for (int i = 0; i < N; i ++)            for (int j = 0; j < N; j ++)                c.a[i][j] = (a[i][j] + rhs.a[i][j]) % mod;        return c;    }};Mat E(2), F(2);void init() {    E.a[0][0] = E.a[1][1] = 1;    F.a[0][0] = F.a[0][1] = F.a[1][0] = 1;}Mat qpow(Mat A, LL b) {    Mat c = E;    while (b > 0) {        if (b & 1) c = A * c;        b >>= 1;        A = A * A;    }    return c;}Mat seg[maxn << 2], col[maxn << 2];bool add[maxn << 2];void PushUp(int v) {    seg[v] = seg[slch] + seg[srch];}void PushDown(int v, int m) {    if (!add[v]) return;    col[slch] = col[slch] * col[v];    col[srch] = col[srch] * col[v];    seg[slch] = seg[slch] * col[v];    seg[srch] = seg[srch] * col[v];    add[slch] = add[srch] = true;    col[v] = E;    add[v] = false;}void build(int l, int r, int v) {    col[v] = E;    if (l == r) {        int k;        scanf("%d", &k);        seg[v] = qpow(F, k - 1);    }    else {        sgetmid;        build(lson);        build(rson);        PushUp(v);    }}void update(int L, int R, int x, int l, int r, int v) {    if (L <= l && r <= R) {        Mat temp = qpow(F, x);         seg[v] = seg[v] * temp;        col[v] = col[v] * temp;        add[v] = true;        return;    }    sgetmid;    PushDown(v, r - l + 1);    if (L <= m) update(L, R, x, lson);    if (R >  m) update(L, R, x, rson);    PushUp(v);}LL query(int L, int R, int l, int r, int v) {    if (L <= l && r <= R) return seg[v].a[0][0];    sgetmid;    PushDown(v, r - l + 1);    LL ans = 0;    if (L <= m) ans = (ans + query(L, R, lson)) % mod;    if (R >  m) ans = (ans + query(L, R, rson)) % mod;    return ans;}int main(int argc, char const *argv[]) {    cin>>N>>M; init(); mem(add, false);    build(1, N, 1);    while (M --) {        int op;        scanf("%d", &op);        if (op == 1) {            int l, r, x;            scanf("%d%d%d", &l, &r, &x);            update(l, r, x, 1, N, 1);        }        else {            int l, r;            scanf("%d%d", &l, &r);            printf("%I64d\n", query(l, r, 1, N, 1));        }    }    return 0;}