Commit 5090d09e authored by 24OI-bot's avatar 24OI-bot
Browse files

style: format markdown files with remark-lint

parent d6db3060

docs/ds/decompose.md

+9 −16
Original line number Diff line number Diff line
@@ -58,31 +58,23 @@ $$ 
    void add(int l, int r, long long x) {

      int sid = id[l], eid = id[r];

      if (sid == eid) {

        for (int i = l; i <= r; i++)

          a[i] += x, s[sid] += x;

        for (int i = l; i <= r; i++) a[i] += x, s[sid] += x;

        return;

      }

      for (int i = l; id[i] == sid; i++)

        a[i] += x, s[sid] += x;

      for (int i = sid + 1; i < eid; i++)

        b[i] += x, s[i] += len * x;

      for (int i = r; id[i] == eid; i--)

        a[i] += x, s[eid] += x;

      for (int i = l; id[i] == sid; i++) a[i] += x, s[sid] += x;

      for (int i = sid + 1; i < eid; i++) b[i] += x, s[i] += len * x;

      for (int i = r; id[i] == eid; i--) a[i] += x, s[eid] += x;

    }

    long long query(int l, int r, long long p) {

      int sid = id[l], eid = id[r];

      long long ans = 0;

      if (sid == eid) {

        for (int i = l; i <= r; i++)

          ans = (ans + a[i] + b[sid]) % p;

        for (int i = l; i <= r; i++) ans = (ans + a[i] + b[sid]) % p;

        return ans;

      }

      for (int i = l; id[i] == sid; i++)

        ans = (ans + a[i] + b[sid]) % p;

      for (int i = sid + 1; i < eid; i++)

        ans = (ans + s[i]) % p;

      for (int i = r; id[i] == eid; i--)

        ans = (ans + a[i] + b[eid]) % p;

      for (int i = l; id[i] == sid; i++) ans = (ans + a[i] + b[sid]) % p;

      for (int i = sid + 1; i < eid; i++) ans = (ans + s[i]) % p;

      for (int i = r; id[i] == eid; i--) ans = (ans + a[i] + b[eid]) % p;

      return ans;

    }

    int main() {
@@ -105,6 +97,7 @@ $$ 
      return 0;

    }

    ```



## 区间和 2



上一个做法的复杂度是 $\Omega(1) , O(\sqrt{n})$ 。