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solutions/P16060.md

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+# P16060 [CSPro 24] 序列查询新解 题解
+
+## 75 pts 暴力
+
+直接遵循题意使用 `lower_bound` 即可得到 75 pts
+
+::::info[[75 pts 代码](https://www.luogu.com.cn/record/303048871)]
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int n, N;
+int a[100005];
+int main() {
+    scanf("%d%d", &n, &N);
+    for (int i = 1; i <= n; ++i) {
+        scanf("%d", &a[i]);
+    }
+    long long ans = 0;
+    for (int i = 0; i <= N - 1; i++) {
+        int f = lower_bound(a, a + n + 1, i) - a;
+        while (a[f] <= i && f <= n)
+            f++;
+        f--;
+        int g = i / (N / (n + 1));
+        ans += abs(f - g);
+    }
+    printf("%lld\n", ans);
+    return 0;
+}
+```
+
+::::
+
+## 100 pts 正解
+
+注意到对于两个数 $a_i, a_{i + 1}$，$f(x) = i, x \in [a_i, a_{i + 1})$
+
+且对于 $g$，只有 $i = k \times\frac{N}{(n + 1)}$ 的时候数值才会变化。
+
+因此，统计 $f, g$ 分别变化的节点进行计算即可。
+
+::::info[[100 pts 代码](https://www.luogu.com.cn/record/303064173)]
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int n, N;
+map<int, int> changes;
+int main() {
+    scanf("%d%d", &n, &N);
+    for (int i = 1, a; i <= n; ++i)
+        scanf("%d", &a), changes[a] = 1;
+    for (int i = 1; i * (N / (n + 1)) < N; i++) {
+        if (changes.find(i * (N / (n + 1))) != changes.end())
+            changes[i * (N / (n + 1))] = 3;
+        else
+            changes[i * (N / (n + 1))] = 2;
+    }
+    changes[N] = 2;
+    int nowf = 0, nowg = 0;
+    long long ans = 0;
+    pair<int, int> pre = {0, 0};
+    for (pair<int, int> now : changes) {
+        ans += abs(nowg - nowf) * (now.first - pre.first);
+        if (now.second == 1)
+            nowf++;
+        else if (now.second == 2)
+            nowg++;
+        else
+            nowf++, nowg++;
+        pre = now;
+    }
+    printf("%lld\n", ans);
+    return 0;
+}
+```
+
+::::