luogu/solutions/P16060.md

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]

#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]

#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;
}

::::