题目描述
思路
代码
#include <cstdio>
#define lc k<<1
#define rc k<<1|1
using namespace std;
int n, m, w;
long long sum[100005 << 2], at[100005];
long long add[100005 << 2], mul[100005 << 2];
inline int read() {
int s = 0;
char ch = getchar();
while (ch < '0' || ch > '9') ch = getchar();
while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s;
}
void build(int k, int l, int r) {
add[k] = 0, mul[k] = 1;
if (l == r) {
sum[k] = at[l] % m;
return;
}
int mid = l + r >> 1;
build(lc, l, mid);
build(rc, mid + 1, r);
sum[k] = (sum[lc] + sum[rc]) % m;
}
void pushdown(int k, int l, int r) {
if (mul[k] != 1) {
mul[lc] = (mul[lc] * mul[k]) % m;
add[lc] = (add[lc] * mul[k]) % m;
sum[lc] = (sum[lc] * mul[k]) % m;
mul[rc] = (mul[rc] * mul[k]) % m;
add[rc] = (add[rc] * mul[k]) % m;
sum[rc] = (sum[rc] * mul[k]) % m;
mul[k] = 1;
}
int mid = l + r >> 1;
if (add[k] != 0) {
add[lc] = (add[lc] + add[k]) % m;
sum[lc] = (sum[lc] + (mid - l + 1) * add[k]) % m;
add[rc] = (add[rc] + add[k]) % m;
sum[rc] = (sum[rc] + (r - mid) * add[k]) % m;
add[k] = 0;
}
}
void upmul(int k, int l, int r, int x, int y, int z) {
if (x <= l && r <= y) {
sum[k] = (sum[k] * z) % m;
mul[k] = (mul[k] * z) % m;
add[k] = (add[k] * z) % m;
return;
}
pushdown(k, l, r);
int mid = l + r >> 1;
if (x <= mid) upmul(lc, l, mid, x, y, z);
if (y > mid) upmul(rc, mid + 1, r, x, y, z);
sum[k] = (sum[lc] + sum[rc]) % m;
}
void upadd(int k, int l, int r, int x, int y, int z) {
if (x <= l && r <= y) {
sum[k] = (sum[k] + (r - l + 1) * z) % m;
add[k] = (add[k] + z) % m;
return;
}
pushdown(k, l, r);
int mid = l + r >> 1;
if (x <= mid) upadd(lc, l, mid, x, y, z);
if (y > mid) upadd(rc, mid + 1, r, x, y, z);
sum[k] = (sum[lc] + sum[rc]) % m;
}
long long query(int k, int l, int r, int x, int y) {
if (x <= l && r <= y) return sum[k] % m;
int mid = l + r >> 1, res = 0;
pushdown(k, l, r);
if (x <= mid) res = query(lc, l, mid, x, y) % m;
if (y > mid) res = (res + query(rc, mid + 1, r, x, y)) % m;
return res % m;
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; ++i) at[i] = read();
build(1, 1, n);
w = read();
for (int i = 1, a, b, c, d, e; i <= w; ++i) {
a = read();
if (a == 1 || a == 2) {
b = read(), c = read(), d = read();
if (a == 1) upmul(1, 1, n, b, c, d % m);
else upadd(1, 1, n, b, c, d % m);
} else {
b = read(), c = read();
printf("%lld
", query(1, 1, n, b, c));
}
}
return 0;
}
这是用queue保存的tag，容易超时
#include <cstdio>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
int n, m, w;
int arr[100005 << 2], at[100005];
queue<pair<int, int> > q[100005 << 2];
inline int read() {
int s = 0;
char ch = getchar();
while (ch < '0' || ch > '9') ch = getchar();
while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s;
}
inline void build(int k, int l, int r) {
if (l == r) {
arr[k] = at[l];
return;
}
int mid = l + r >> 1;
build(k << 1, l, mid);
build(k << 1 | 1, mid + 1, r);
arr[k] = (0LL + arr[k << 1] + arr[k << 1 | 1]) % m;
}
inline void trans(int k, int l, int r, pair<int, int> p) {
q[k].push(p);
if (p.first == 1)
arr[k] = 1LL * arr[k] * p.second % m;
else
arr[k] = (0LL + arr[k] + (r - l + 1) * p.second) % m;
}
inline void pushdown(int k, int l, int r) {
int L = k << 1, R = k << 1 | 1, mid = l + r >> 1;
while (!q[k].empty()) {
trans(L, l, mid, q[k].front());
trans(R, mid + 1, r, q[k].front());
q[k].pop();
}
}
inline void change(int k, int l, int r, int x, int y, int p, int q) {
if (x <= l && r <= y) {
trans(k, l, r, make_pair(p, q));
return;
}
pushdown(k, l, r);
int mid = l + r >> 1;
if (x <= mid)
change(k << 1, l, mid, x, y, p, q);
if (y > mid)
change(k << 1 | 1, mid + 1, r, x, y, p, q);
arr[k] = (0LL + arr[k << 1] + arr[k << 1 | 1]) % m;
}
inline int query(int k, int l, int r, int x, int y) {
if (x <= l && r <= y)
return arr[k];
int mid = l + r >> 1;
int res = 0;
pushdown(k, l, r);
if (x <= mid)
res = query(k << 1, l, mid, x, y);
if (y > mid)
res = (0LL + res + query(k << 1 | 1, mid + 1, r, x, y)) % m;
return res;
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; ++i) at[i] = read();
build(1, 1, n);
w = read();
for (int i = 1, a, b, c, d, e; i <= w; ++i) {
a = read();
if (a == 1 || a == 2) {
b = read(), c = read(), d = read();
if (a == 1)
change(1, 1, n, b, c, 1, d);
else
change(1, 1, n, b, c, 2, d);
} else {
b = read(), c = read();
printf("%d
", query(1, 1, n, b, c));
}
}
return 0;
}