大部分题解楼上的巨巨已经给出了，这里补充下I，K，L的题解

I 最大公约数（简单数论）

思路：

对任意x，若x为序列的因数，即有:

$x\mid a[1]x\backslash mid\; a[1]$ ,$~~~$ $x\mid a[2]x\backslash mid\; a[2]$,$~~~$ $x\mid a[3]x\backslash mid\; a[3]$ $~~~$ .....$~~~$ $x\mid a[n]x\backslash mid\; a[n]$.$~~~$ 那么令$sum=a[1]+a[2]+\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}+a[n]sum=a[1]+a[2]+....+a[n]$; 显然有$x\mid sumx\backslash mid\; sum$

即最多可能产生的序列的因数的数量即为$sumsum$的不同因数个数。直接对序列求和，计算和的不同因数的个数即可。

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
const double eps = 1e-9;
void solve(ll sum)
{
    ll ans = 0;
    for (ll i = 1; i <= sum / i; i++)
        if (sum % i == 0)
            if (i * i == sum)
                ans++;
            else
                ans += 2;
    cout << ans << endl;
}
int main()
{
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

#ifdef ak
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int n;
    cin >> n;
    ll sum = 0;
    for (int i = 1; i <= n; i++)
    {
        ll t;
        cin >> t;
        sum += t;
    }
    solve(sum);
    return 0;
}

K 金牌厨师（二分答案+差分维护）

思路：

二分最大化可行答案，对于每次mid，差分维护判断是否存在连续长度不小于mid且人数也不小于mid的区间即可。

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
const double eps = 1e-9;
const int N = 3e5 + 10;
pll a[N];
int sum[N], n, m;
bool check(int mid)
{
    memset(sum, 0, sizeof sum);
    for (int i = 1; i <= m; i++)
    {
        int len = a[i].second - a[i].first + 1;
        if (len >= mid)
        {
            sum[a[i].first + mid - 1]++;
            sum[a[i].second + 1]--;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        sum[i] += sum[i - 1];
        if (sum[i] >= mid)
            return 1;
    }
    return 0;
}

int main()
{
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

#ifdef ak
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
        cin >> a[i].first >> a[i].second;

    int l = 1, r = n;
    while (l < r)
    {
        int mid = (l + r + 1) >> 1;
        if (check(mid))
            l = mid;
        else
            r = mid - 1;
    }
    cout << l << endl;

    return 0;
}

L WireConnection（最小生成树裸题）

思路：

三维上求最小生成树即可,这里给出prim算法实现

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
const double eps = 1e-9;
const int N = 1e4 + 10;
ll dis[N], ans = 0, n, st[N];
struct electric
{
    ll x, y, z;
} el[N];
ll calcu(int a, int b)
{
    ll w = ceil(sqrt((el[a].x - el[b].x) * (el[a].x - el[b].x) + (el[a].y - el[b].y) * (el[a].y - el[b].y) + (el[a].z - el[b].z) * (el[a].z - el[b].z)));
    return w;
}
void prim()
{
    memset(dis, 0x3f, sizeof dis);
    dis[1] = 0;
    for (int i = 1; i <= n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dis[t] > dis[j]))
                t = j;

        ans += dis[t];
        st[t] = 1;

        for (int j = 1; j <= n; j++)
            if (!st[j])
                dis[j] = min(dis[j], calcu(j, t));
    }
}
int main()
{
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

#ifdef ak
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif

    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        ll a, b, c;
        cin >> a >> b >> c;
        el[i] = {a, b, c};
    }
    prim();
    cout << ans << endl;
    return 0;
}