bzoj 2962 序列操作(线段树)

题外话

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做这道题我整个人都非常的绝望，推了一会发现是线段树裸题，然后调了N久一直是WA

情况是这样的

开始WA的几百毫秒的都是由于我比较SB造成的，可是跑了10几秒的程序我查了N久也查不出错

最后灵机一动把50000改成60000就过了，也不知道为啥T_T

Description

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一个长度为n的序列，有3种操作

1:区间加c

2:区间取为相反数

3:询问区间中选择c个数相乘的所有方案的和mod19940417的值

Solution

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这个操作3非常鬼畜，似乎没啥好的办法，但是仔细推导一番会发现这个东西其实是可以区间合并的

用f[i]表示选i个数所有方案和，rt.f[i]=∑ij=0ls.f[j]∗rs.f[i−j]

然后注意一点f[0]=1

对于第一种操作加k我们发现对答案f[i]会变为∑ij=0f[j]∗Cin−j∗ki−j(想一想为什么)

对于第二种操作取反，我们发现只有当c为奇数时答案会取反，这个也很好搞

然后就是打两个标记reverse和add的线段树辣，注意处理标记的顺序

Code

#include <bits/stdc++.h>using namespace std;typedef long long LL;#define ls (rt << 1)#define rs (rt << 1 | 1)const int N = 60050, M = 19940417;int a[N], b[25], c[N][25], add[N << 2], rev[N << 2];struct node {    int f[25];}p[N << 2];inline int read(int &t) {    int f = 1;char c;    while (c = getchar(), c < '0' || c > '9') if (c == '-') f = -1;    t = c - '0';    while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0';    t *= f;}inline void U(int &x, int y) {    x += y;    if (x >= M) x -= M;    if (x < 0)  x += M;}void calc(int rt, int l, int r, int x) {    for (int i = 1; i <= 20; ++i) {        int t = 0, tx = 1;        for (int j = i; j >= 0; --j) {            U(t, (LL)p[rt].f[j] * c[r - l + 1 - j][i - j] % M * tx % M);            tx = (LL)tx * x % M;        }        b[i] = t;    }       for (int i = 1; i <= 20; ++i)   p[rt].f[i] = b[i];}void down(int rt, int l, int r) {    if (rev[rt]) {        rev[ls] ^= 1, rev[rs] ^= 1;        for (int i = 1; i <= 20; i += 2)    p[ls].f[i] = (M - p[ls].f[i]) % M;        for (int i = 1; i <= 20; i += 2)    p[rs].f[i] = (M - p[rs].f[i]) % M;        add[ls] = (M - add[ls]) % M, add[rs] = (M - add[rs]) % M;        rev[rt] = 0;    }    if (add[rt]) {        int mid = l + r >> 1;        U(add[ls], add[rt]), U(add[rs], add[rt]);        calc(ls, l, mid, add[rt]), calc(rs, mid + 1, r, add[rt]);        add[rt] = 0;    }} void up(int rt) {    for (int i = 1; i <= 20; ++i)   p[rt].f[i] = 0;    for (int i = 0; i <= 20; ++i)           for (int j = 0; i + j <= 20; ++j) {            if (i + j == 0) continue;            U(p[rt].f[i + j], (LL)p[ls].f[i] * p[rs].f[j] % M);        }}void build(int rt, int l, int r) {    p[rt].f[0] = 1;    if (l == r) {        p[rt].f[1] = a[l];        return;    }    int mid = l + r >> 1;    build(ls, l, mid), build(rs, mid + 1, r);    up(rt);}void reverse(int rt, int l, int r, int L, int R) {    if (L <= l && R >= r) {        rev[rt] ^= 1;        add[rt] = (M - add[rt]) % M;        for (int i = 1; i <= 20; i += 2) p[rt].f[i] = (M - p[rt].f[i]) % M;        return;    }    down(rt, l, r);    int mid = l + r >> 1;    if (L <= mid)   reverse(ls, l, mid, L, R);    if (R > mid)    reverse(rs, mid + 1, r, L, R);    up(rt);}void inc(int rt, int l, int r, int L, int R, int x) {    if (L <= l && R >= r) {        U(add[rt], x);        calc(rt, l, r, x);        return;    }    int mid = l + r >> 1;    down(rt, l, r);    if (L <= mid)   inc(ls, l, mid, L, R, x);    if (R > mid)    inc(rs, mid + 1, r, L, R, x);    up(rt);}node ask(int rt, int l, int r, int L, int R) {    if (L <= l && R >= r) {        return p[rt];    }    down(rt, l, r);    int mid = l + r >> 1;    node t, t1, t2;    if (L > mid)    t = ask(rs, mid + 1, r, L, R);    else if (R <= mid)  t = ask(ls, l, mid, L, R);    else {        t.f[0] = 1;        for (int i = 1; i <= 20; ++i)   t.f[i] = 0;        t1 = ask(ls, l, mid, L, mid), t2 = ask(rs, mid + 1, r, mid + 1, R);        for (int i = 0; i <= 20; ++i)            for (int j = 0; j + i <= 20; ++j) {                if (i + j == 0) continue;                U(t.f[i + j], (LL)t1.f[i] * t2.f[j] % M);            }    }       return t;}   int main() {    int n, q;    read(n), read(q);    for (int i = 0; i <= n; ++i) {        c[i][0] = c[i][i] = 1;        for (int j = 1; j < i; ++j) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % M;    }    for (int i = 1; i <= n; ++i)    read(a[i]), a[i] = (a[i] % M + M) % M;    build(1, 1, n);    char s[5];    int x, y, z;    while (q--) {        scanf("%s", s);        if (s[0] == 'I') {            read(x), read(y), read(z);            z = (z % M + M) % M;            inc(1, 1, n, x, y, z);        }        else if (s[0] == 'R') {            read(x), read(y);            reverse(1, 1, n, x, y);        }        else {            read(x), read(y), read(z);            z = (z % M + M) % M;            node t = ask(1, 1, n, x, y);            printf("%d\n", t.f[z] % M);        }    }    return 0;}