112 lines
2.8 KiB
C++
112 lines
2.8 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1000010;
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int n, m, p;
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// 线段树模板
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#define LL long long
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#define ls u << 1
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#define rs u << 1 | 1
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#define mid ((l + r) >> 1)
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struct Node {
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int l, r;
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int mu, add;
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LL sum;
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} tr[N << 2];
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void pushup(int u) {
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tr[u].sum = (tr[ls].sum + tr[rs].sum) % p;
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}
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void build(int u, int l, int r) {
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tr[u].l = l, tr[u].r = r, tr[u].mu = 1;
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if (l == r) {
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cin >> tr[u].sum;
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return;
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}
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build(ls, l, mid), build(rs, mid + 1, r);
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pushup(u);
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}
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void pushdown(int u) {
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if (tr[u].add == 0 && tr[u].mu == 1) return; // 默认懒标记
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int &mu = tr[u].mu, &add = tr[u].add; // 此时的add懒标记，已经处理过了
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tr[ls].sum = ((LL)mu * tr[ls].sum % p + (LL)(tr[ls].r - tr[ls].l + 1) * add % p) % p;
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tr[rs].sum = ((LL)mu * tr[rs].sum % p + (LL)(tr[rs].r - tr[rs].l + 1) * add % p) % p;
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tr[ls].mu = (LL)tr[ls].mu * mu % p;
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tr[rs].mu = (LL)tr[rs].mu * mu % p;
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tr[ls].add = ((LL)tr[ls].add * mu % p + add) % p;
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tr[rs].add = ((LL)tr[rs].add * mu % p + add) % p;
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mu = 1, add = 0; // 清空懒标记
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}
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void add(int u, int L, int R, int v) {
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int l = tr[u].l, r = tr[u].r;
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if (l >= L && r <= R) {
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tr[u].add = ((LL)tr[u].add + v) % p;
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tr[u].sum = ((LL)tr[u].sum + v * ((LL)tr[u].r - tr[u].l + 1) % p) % p;
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return;
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}
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if (l > R || r < L) return;
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pushdown(u);
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add(ls, L, R, v), add(rs, L, R, v);
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pushup(u);
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}
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void mu(int u, int L, int R, int v) {
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int l = tr[u].l, r = tr[u].r;
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if (l >= L && r <= R) {
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tr[u].add = (LL)tr[u].add * v % p; // 比较重要的一步,add要在这里乘上v,因为后面可能要加其他的数而那些数其实是不用乘k的
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tr[u].mu = (LL)tr[u].mu * v % p;
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tr[u].sum = (LL)tr[u].sum * v % p;
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return;
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}
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if (l > R || r < L) return;
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pushdown(u);
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mu(ls, L, R, v), mu(rs, L, R, v);
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pushup(u);
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}
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LL query(int u, int L, int R) {
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int l = tr[u].l, r = tr[u].r;
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if (l >= L && r <= R) return tr[u].sum;
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if (l > R || r < L) return 0;
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pushdown(u);
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return (query(ls, L, R) + query(rs, L, R)) % p;
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}
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signed main() {
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// 文件输入输出
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#ifndef ONLINE_JUDGE
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freopen("P3373.in", "r", stdin);
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#endif
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// 加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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cin >> n >> m >> p;
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build(1, 1, n);
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while (m--) {
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int op, l, r;
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cin >> op >> l >> r;
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if (op == 1) {
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int k;
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cin >> k;
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mu(1, l, r, k);
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} else if (op == 2) {
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int k;
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cin >> k;
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add(1, l, r, k);
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} else
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printf("%lld\n", query(1, l, r));
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}
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return 0;
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} |