#6280. 数列分块入门 4

https://loj.ac/s/1465673

LibreOJ/6280/6280.cpp

@@ -0,0 +1,92 @@
+#include <cmath>
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 50005;
+
+int n, t, c, a[N], st[N], ed[N], pos[N];
+long long sum[N], add[N];
+
+void change(int l, int r, int c) {
+    int p = pos[l],
+        q = pos[r];
+
+    if (p == q) {
+        for (int i = l; i <= r; i++) {
+            a[i] += c;
+        }
+        sum[p] += (r - l + 1) * c;
+
+        return;
+    }
+
+    for (int i = l; i <= ed[p]; i++) a[i] += c;
+    sum[p] += (ed[p] - l + 1) * c;
+
+    for (int i = st[q]; i <= r; i++) a[i] += c;
+    sum[q] += (r - st[q] + 1) * c;
+
+    for (int i = p + 1; i <= q - 1; i++) add[i] += c;
+}
+
+int query(int l, int r, int c) {
+    int p = pos[l],
+        q = pos[r];
+    int res = 0;
+
+    if (p == q) {
+        for (int i = l; i <= r; i++) {
+            res = (res + a[i] + add[p]) % (c + 1);
+        }
+
+        return res;
+    }
+
+    for (int i = l; i <= ed[p]; i++) res = (res + a[i] + add[p]) % (c + 1);
+    for (int i = st[q]; i <= r; i++) res = (res + a[i] + add[q]) % (c + 1);
+
+    for (int i = p + 1; i <= q - 1; i++) res = (res + sum[i] + 1ll * add[i] * t) % (c + 1);
+
+    return res;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+
+    cin >> n;
+
+    for (int i = 1; i <= n; i++) {
+        cin >> a[i];
+    }
+
+    t = std::sqrt(n);
+
+    for (c = 1;; c++) {
+        st[c] = (c - 1) * t + 1;
+        ed[c] = std::min(n, t * c);
+
+        if (t * c >= n) break;
+    }
+
+    for (int i = 1; i <= c; i++) {
+        for (int j = st[i]; j <= ed[i]; j++) {
+            pos[j] = i;
+            sum[i] += a[j];
+        }
+    }
+
+    for (int i = 1, op, l, r, c; i <= n; i++) {
+        cin >> op >> l >> r >> c;
+
+        if (op == 0) {
+            change(l, r, c);
+        } else {  // op == 1
+            cout << query(l, r, c) % (c + 1) << endl;
+        }
+    }
+
+    return 0;
+}