题外话
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
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
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;
}
时间： 2024-10-21 22:17:38