【luogu CF1109E】Sasha and a Very Easy Test（线段树）
Sasha and a Very Easy Test
题目链接：luogu CF1109E
题目大意
维护一个长度为 n 的序列，有区间乘，单点除（保证能整除），区间求和答案对 p 取模。
p 不一定是质数。
思路
麻了考场被卡了常，重构了一遍之后发现改时限了，考场的代码能过了。
难泵。
首先如果 \(p\) 是质数，那直接上逆元乱搞都行。（线段树）
但是不是质数，但是我们考虑直接暴力拆解，因为有问题的时候是出现倍数。
那拆出来的质数，我们维护的值就可以用一个余数和一堆次数表示。
分别是跟 \(p\) 互质的部分，以及带上了多少个某个质数的次数。
然后维护就好了。
一个小小注意的点是我们这种分解只需要在最下层进行，因为我们除法是单点的，而且是要区间求和。
所以上面的我们直接拿值来加就可以，维护最下层的 \(n\) 个即可。
代码
考场代码（比较慢，被考场的 3s 卡了时间，虽然 luogu 上很快而且考场后改成了 8s 之后我跑了 4s
#include<map>
#include<cstdio>
#include<vector>
using namespace std;
const int N = 5e5 + 100;
int n, p, q, a[N], zs[35], tot;
map <int, int> pla;
vector <int> mic[35]; int R[35];
int add(int x, int y) {return x + y >= p ? x + y - p : x + y;}
int dec(int x, int y) {return x < y ? x - y + p : x - y;}
int mul(int x, int y) {return 1ll * x * y % p;}
int re, zf; char c;
int read() {
re = 0; zf = 1; c = getchar();
while (c < '0' || c > '9') {
if (c == '-') zf = -zf;
c = getchar();
}
while (c >= '0' && c <= '9') {
re = (re << 3) + (re << 1) + c - '0';
c = getchar();
}
return re * zf;
}
struct Val {
int x, a[35];
};
Val operator *(Val x, Val y) {
Val re = x; re.x = mul(re.x, y.x);
for (int i = 0; i < tot; i++) re.a[i] += y.a[i];
return re;
}
int exgcd(int a, int b, int &x, int &y) {
if (!b) {
x = 1; y = 1; return a;
}
int re = exgcd(b, a % b, y, x);
y -= a / b * x; return re;
}
int get_inv(int now, int p) {
int x, y; exgcd(now, p, x, y);
return (x % p + p) % p;
}
Val operator /(Val x, Val y) {
Val re = x; re.x = mul(re.x, get_inv(y.x, p));
for (int i = 0; i < tot; i++) re.a[i] -= y.a[i];
return re;
}
Val getVAL(int x) {
Val re;
re.x = x;
for (int i = 0; i < tot; i++) {
re.a[i] = 0;
while (re.x % zs[i] == 0) re.a[i]++, re.x /= zs[i];
}
return re;
}
int msm(int id, int k) {//可以直接记忆化预处理次方，就不用快速幂，因为至多是大概 30*(m+1) 大概是 1e7 级别的
while (k > R[id]) {
mic[id].push_back(mul(mic[id][R[id]], zs[id])); R[id]++;
}
return mic[id][k];
}
int getval(Val x) {
int re = x.x;
for (int i = 0; i < tot; i++)
re = mul(re, msm(i, x.a[i]));
return re;
}
bool check_empty(Val x) {
if (x.x != 1) return 0;
for (int i = 0; i < tot; i++) if (x.a[i]) return 0;
return 1;
}
struct XD_Tree {
Val lzy[N << 2], val[N];
int sum[N << 2], leaf[N << 2];
void up(int now) {
sum[now] = add(sum[now << 1], sum[now << 1 | 1]);
}
void downm(int now, Val a) {
sum[now] = mul(sum[now], getval(a));
lzy[now] = lzy[now] * a;
if (leaf[now]) val[leaf[now]] = val[leaf[now]] * a;
}
void down(int now) {
if (check_empty(lzy[now])) return ;
downm(now << 1, lzy[now]); downm(now << 1 | 1, lzy[now]);
lzy[now].x = 1; for (int i = 0; i < tot; i++) lzy[now].a[i] = 0;
}
void build(int now, int l, int r) {
lzy[now].x = 1; for (int i = 0; i < tot; i++) lzy[now].a[i] = 0;
if (l == r) {
leaf[now] = l;
val[l] = getVAL(a[l]); sum[now] = a[l] % p; return ;
}
int mid = (l + r) >> 1;
build(now << 1, l, mid); build(now << 1 | 1, mid + 1, r);
up(now);
}
int query(int now, int l, int r, int L, int R) {
if (L <= l && r <= R) return sum[now];
down(now); int mid = (l + r) >> 1, re = 0;
if (L <= mid) re = add(re, query(now << 1, l, mid, L, R));
if (mid < R) re = add(re, query(now << 1 | 1, mid + 1, r, L, R));
return re;
}
void times(int now, int l, int r, int L, int R, Val a) {
if (L <= l && r <= R) {
downm(now, a); return ;
}
down(now); int mid = (l + r) >> 1;
if (L <= mid) times(now << 1, l, mid, L, R, a);
if (mid < R) times(now << 1 | 1, mid + 1, r, L, R, a);
up(now);
}
void inv(int now, int l, int r, int pl, Val x) {
if (l == r) {
val[l] = val[l] / x; sum[now] = getval(val[l]);
return ;
}
down(now); int mid = (l + r) >> 1;
if (pl <= mid) inv(now << 1, l, mid, pl, x);
else inv(now << 1 | 1, mid + 1, r, pl, x);
up(now);
}
}T;
int main() {
n = read(); p = read();
int tmp = p;
for (int i = 2; i * i <= tmp; i++)
if (tmp % i == 0) {
zs[tot++] = i; pla[i] = tot - 1;
while (tmp % i == 0) tmp /= i;
}
if (tmp > 1) zs[tot++] = tmp, pla[tmp] = tot - 1;
for (int i = 0; i < tot; i++) mic[i].push_back(1);
for (int i = 1; i <= n; i++) a[i] = read();
T.build(1, 1, n);
q = read();
while (q--) {
int op = read();
if (op == 1) {
int l = read(), r = read(), x = read();
T.times(1, 1, n, l, r, getVAL(x));
}
if (op == 2) {
int P = read(), x = read();
T.inv(1, 1, n, P, getVAL(x));
}
if (op == 3) {
int l = read(), r = read();
printf("%d\n", T.query(1, 1, n, l, r));
}
// printf("%d\n", T.val[5].x);
// for (int i = 0; i < tot; i++) printf("%d ", T.val[5].a[i]);
// printf("\n");
}
return 0;
}
改进代码（根据别人的写法改进了一下，考后测只需要 1s 左右
#include<map>
#include<cstdio>
#include<vector>
using namespace std;
const int N = 5e5 + 100;
int n, p, q, a[N], phi, prime[10], tot;
int tmp[10];
int add(int x, int y) {return x + y >= p ? x + y - p : x + y;}
int dec(int x, int y) {return x < y ? x - y + p : x - y;}
int mul(int x, int y) {return 1ll * x * y % p;}
int re, zf; char c;
int read() {
re = 0; zf = 1; c = getchar();
while (c < '0' || c > '9') {
if (c == '-') zf = -zf;
c = getchar();
}
while (c >= '0' && c <= '9') {
re = (re << 3) + (re << 1) + c - '0';
c = getchar();
}
return re * zf;
}
int get_phi(int n) {
int ans = n;
for (int i = 2; i * i <= n; i++)
if (n % i == 0) {
ans = ans / i * (i - 1);
while (n % i == 0) n /= i;
}
if (n > 1) ans = ans / n * (n - 1);
return ans;
}
void Init() {
phi = get_phi(p);
int tmp = p;
for (int i = 2; i * i <= tmp; i++)
if (tmp % i == 0) {
prime[++tot] = i; while (tmp % i == 0) tmp /= i;
}
if (tmp > 1) prime[++tot] = tmp;
}
int ksm(int x, int y) {
int re = 1;
while (y) {
if (y & 1) re = mul(re, x);
x = mul(x, x); y >>= 1;
}
return re;
}
int work(int n) {
for (int i = 1; i <= tot; i++) {
tmp[i] = 0; while (n % prime[i] == 0) n /= prime[i], tmp[i]++;
}
return n % p;
}
struct XD_tree {
int sum[N << 2], lzy[N << 2], res[N << 2], s[N << 2][10];
void up(int now) {
sum[now] = add(sum[now << 1], sum[now << 1 | 1]);
}
void downm(int now, int val, int re, int *tmp) {
lzy[now] = mul(lzy[now], val); sum[now] = mul(sum[now], val);
res[now] = mul(res[now], re);
for (int i = 1; i <= tot; i++) s[now][i] += tmp[i];
}
void down(int now) {
if (lzy[now] != 1 || res[now] != 1) {
downm(now << 1, lzy[now], res[now], s[now]);
downm(now << 1 | 1, lzy[now], res[now], s[now]);
lzy[now] = res[now] = 1;
for (int i = 1; i <= tot; i++) s[now][i] = 0;
}
}
void build(int now, int l, int r) {
lzy[now] = res[now] = 1;
if (l == r) {
sum[now] = a[l] % p;
res[now] = work(a[l]);
for (int i = 1; i <= tot; i++) s[now][i] = tmp[i];
return ;
}
int mid = (l + r) >> 1;
build(now << 1, l, mid); build(now << 1 | 1, mid + 1, r);
up(now);
}
void times(int now, int l, int r, int L, int R, int val, int re) {
if (L <= l && r <= R) {
downm(now, val, re, tmp);
return ;
}
down(now); int mid = (l + r) >> 1;
if (L <= mid) times(now << 1, l, mid, L, R, val, re);
if (mid < R) times(now << 1 | 1, mid + 1, r, L, R, val, re);
up(now);
}
void inv(int now, int l, int r, int pl, int val) {
if (l == r) {
res[now] = mul(res[now], ksm(val, phi - 1));
sum[now] = res[now];
for (int i = 1; i <= tot; i++) s[now][i] -= tmp[i], sum[now] = mul(sum[now], ksm(prime[i], s[now][i]));
return ;
}
down(now); int mid = (l + r) >> 1;
if (pl <= mid) inv(now << 1, l, mid, pl, val);
else inv(now << 1 | 1, mid + 1, r, pl, val);
up(now);
}
int query(int now, int l, int r, int L, int R) {
if (L <= l && r <= R) return sum[now];
down(now); int mid = (l + r) >> 1, re = 0;
if (L <= mid) re = add(re, query(now << 1, l, mid, L, R));
if (mid < R) re = add(re, query(now << 1 | 1, mid + 1, r, L, R));
return re;
}
}T;
int main() {
n = read(); p = read();
for (int i = 1; i <= n; i++) a[i] = read();
Init(); T.build(1, 1, n);
q = read();
while (q--) {
int op = read();
if (op == 1) {
int l = read(), r = read(), x = read();
T.times(1, 1, n, l, r, x, work(x));
}
if (op == 2) {
int pl = read(), x = read();
T.inv(1, 1, n, pl, work(x));
}
if (op == 3) {
int l = read(), r = read();
printf("%d\n", T.query(1, 1, n, l, r));
}
}
return 0;
}
