From ea0e1d3d6ee126a3f8de68e59b38285136591f1e Mon Sep 17 00:00:00 2001
From: Ren Baoshuo <i@baoshuo.ren>
Date: Thu, 26 Nov 2020 19:43:41 +0800
Subject: [PATCH] =?UTF-8?q?P3373=20=E3=80=90=E6=A8=A1=E6=9D=BF=E3=80=91?=
 =?UTF-8?q?=E7=BA=BF=E6=AE=B5=E6=A0=91=202?=
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R42643505
---
 problem/P3373/P3373.cpp | 131 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 131 insertions(+)
 create mode 100644 problem/P3373/P3373.cpp

diff --git a/problem/P3373/P3373.cpp b/problem/P3373/P3373.cpp
new file mode 100644
index 00000000..6677f05e
--- /dev/null
+++ b/problem/P3373/P3373.cpp
@@ -0,0 +1,131 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+struct node {
+    int l, r;
+    long long s, d, x;
+
+    node() {
+        l = r = s = d = 0;
+        x = 1;
+    }
+    node(int _l, int _r) {
+        l = _l, r = _r;
+        s = d = 0;
+        x = 1;
+    }
+} tr[100005 << 2];
+int n, m, p, op, x, y;
+long long k, a[100005];
+
+/**
+ * @brief 上传信息
+ * @param u 父节点下标
+ */
+void pushup(int u) {
+    tr[u].s = (tr[u << 1].s + tr[u << 1 | 1].s) % p;
+}
+
+/**
+ * @brief 下放懒标记
+ * @param u 父节点下标
+ * @attention 先乘后加
+ */
+void pushdown(int u) {
+    // 左子树
+    tr[u << 1].s = ((tr[u << 1].s * tr[u].x) + (tr[u << 1].r - tr[u << 1].l + 1) * tr[u].d) % p;
+    tr[u << 1].x = tr[u << 1].x * tr[u].x % p;
+    tr[u << 1].d = (tr[u << 1].d * tr[u].x + tr[u].d) % p;
+    // 右子树
+    tr[u << 1 | 1].s = ((tr[u << 1 | 1].s * tr[u].x) + (tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1) * tr[u].d) % p;
+    tr[u << 1 | 1].x = tr[u << 1 | 1].x * tr[u].x % p;
+    tr[u << 1 | 1].d = (tr[u << 1 | 1].d * tr[u].x + tr[u].d) % p;
+    // 清除懒标记
+    tr[u].d = 0;
+    tr[u].x = 1;
+}
+
+/**
+ * @brief 建立线段树
+ * @param u 根节点下标
+ * @param l 左端点
+ * @param r 右端点
+ */
+void build(int u, int l, int r) {
+    tr[u] = node(l, r);
+    if (l == r) {
+        tr[u].s = a[l] % p;
+        return;
+    }
+    int mid = l + r >> 1;
+    build(u << 1, l, mid);
+    build(u << 1 | 1, mid + 1, r);
+    pushup(u);
+}
+
+/**
+ * @brief 区间加
+ * @details 将区间 [l, r] 中的每一个数加上 d
+ * @param u 父节点下标
+ * @param l 左端点
+ * @param r 右端点
+ * @param x 乘上的数
+ * @param d 增加的值
+ */
+void modify(int u, int l, int r, long long x, long long d) {
+    // 被包含直接修改
+    if (tr[u].l >= l && tr[u].r <= r) {
+        tr[u].s = ((tr[u].s * x) + (tr[u].r - tr[u].l + 1) * d) % p;
+        tr[u].x = tr[u].x * x % p;
+        tr[u].d = (tr[u].d * x + d) % p;
+        return;
+    }
+    int mid = tr[u].l + tr[u].r >> 1;
+    pushdown(u);                                  // 下放懒标记
+    if (l <= mid) modify(u << 1, l, r, x, d);     // 和左侧有交集
+    if (r > mid) modify(u << 1 | 1, l, r, x, d);  // 和右侧有交集
+    pushup(u);                                    // 上传新信息
+}
+
+/**
+ * @brief 求区间和
+ * @param u 
+ * @param l 
+ * @param r 
+ * @return int 
+ */
+long long query(int u, int l, int r) {
+    if (tr[u].l >= l && tr[u].r <= r) {  // 被包含直接返回
+        return tr[u].s;
+    }
+    int mid = tr[u].l + tr[u].r >> 1;
+    long long s = 0;
+    pushdown(u);                                         // 下放懒标记
+    if (l <= mid) s = query(u << 1, l, r);               // 和左侧有交集
+    if (r > mid) s = (s + query(u << 1 | 1, l, r)) % p;  // 和右侧有交集
+    return s;
+}
+
+signed main() {
+    cin >> n >> m >> p;
+    for (int i = 1; i <= n; i++) {
+        cin >> a[i];
+    }
+    build(1, 1, n);
+    while (m--) {
+        cin >> op >> x >> y;
+        if (op == 1) {
+            cin >> k;
+            modify(1, x, y, k, 0);
+        }
+        else if (op == 2) {
+            cin >> k;
+            modify(1, x, y, 1, k);
+        }
+        else if (op == 3) {
+            cout << query(1, x, y) % p << endl;
+        }
+    }
+    return 0;
+}