牛牛的揠苗助长

一眼二分分天数,然后就是要找合适的高度满足操作数小于天数
两种方法:

二分的check函数里面三分高度,我们可以发现高度关于操作数是一个凹函数.
要注意临界值

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 7;
ll a[maxn], n;
ll cheak3(ll x)
{
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        ans += abs(a[i] - x);
    return ans;
}
int cheak(ll m)
{
    // printf ("**\n");
    for (int i = 1; i <= m % n; i++)
        a[i] += 1;
    ll l = 0, r = 1e9 + 7, mid1, mid2, t = 100, ans = 1e15;
    while (t--)
    {
        mid1 = (l + r) / 2;
        mid2 = (mid1 + r) / 2;
        if (cheak3(mid1) > cheak3(mid2))
            l = mid1;
        else
            r = mid2;
        // printf ("%lld %lld\n",l,r);
    }
    for (int x = -20; x <= 20; x++)
    {
        if (l + x <= 0)
            continue;
        ans = min(ans, cheak3(l + x));
    }
    for (int i = 1; i <= m % n; i++)
        a[i] -= 1;
    return ans <= m;
}
int main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    ll l = 1, r = 1e15, mid, ans = 1;
    //printf ("**\n");
    while (l <= r)
    {
        mid = (l + r) / 2;
        if (cheak(mid))
            ans = mid, r = mid - 1;
        else
            l = mid + 1;
        // printf ("%lld %lld\n",l,r);
    }
    printf("%lld\n", ans);
    return 0;
}

第二种就是对变化后的水稻排序找中位数,合适的高度就是其中位数
推导
a为排序过的数组
从2开始: a1,a2要找一个h使操作数abs(a1-h)+abs(a2-h)=a2-a1,显然在a1,a2间均可;
然后到3: a1,a2,a3,由上面推得h要在a1,a3间,明显a2是最优,操作数是a3-a1;
然后到4: a1,a2,a3,a4,明显,最优h在a2,a3之间 操作数是a4-a1+a3-a2;
当为n时: 抽象在数轴上,即为求解点h使h到数轴上个点距离之和最近,推得是中位数
所以就能愉快的少一次三分hhh

#include <bits/stdc++.h>
#define ll long long
#define ios ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

using namespace std;
const ll inf = 0x3f3f3f3f;
const ll maxn = 1e5 + 7;
ll a[maxn];
ll n;
ll sum = 0;
bool check(ll t)
{
    ll b[maxn];
    ll opsum1 = 0;
    ll add = t / n;
    ll tt = t % n;
    for (ll i = 1; i <= n; i++)
    {
        if (i > tt)
            b[i] = a[i] + add;
        else
            b[i] = a[i] + add + 1;
    }
    sort(b + 1, b + n + 1);
    for (int i = 1; i <= n / 2; i++)
    {
        opsum1 += b[n - i + 1] - b[i];
    }
    return opsum1 <= t;
}
int main()
{
    ios;
    cin >> n;
    for (ll i = 1; i <= n; i++)
    {
        cin >> a[i];
        sum += a[i];
    }
    ll l = 1, r = 1e15, mid, ans = 1;
    //printf ("**\n");
    while (l <= r)
    {
        mid = (l + r) / 2;
        if (check(mid))
            ans = mid, r = mid - 1;
        else
            l = mid + 1;
        // printf ("%lld %lld\n",l,r);
    }
    printf("%lld\n", ans);

    return 0;
}