A - Another String Minimization 问题

水

B - Making Towers

很显然两个间隔为偶数的相同数字可以连在一起

C - Qpwoeirut And The City

奇数显然。偶数可以认为是选择一个偶数前缀和奇数后缀，结束了。

D - Chopping Carrots

不难想到枚举最小值然后二分 check，然后优化就是把最小值整除分块。

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#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int , int > pii;
typedef unsigned long long ull;

namespace FastIO 
{
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
};
using namespace FastIO;

const int N = 1e5+10;

int a[N], te[N], mxn[N];
int n, k;

inline int get(int l, int n) {
    return n / (n / l);
}

inline void Input() {
    read(n, k);
    for(int i = 1; i <= n; i++) {
        read(a[i]);
    }
}

inline void Work() {
    for(int i = 1; i <= n; i++) mxn[i] = 0;
    int mx = 0;
    for(int i = 1; i <= n; i++) {
        int tot = 0;
        for(int l = 1, r = 0; l <= min(k, a[i]); l = r + 1) {
            r = get(l, a[i]);
            te[++tot] = a[i] / l;
        }
        if(k > a[i]) te[++tot] = 0;
        reverse(te + 1, te + 1 + tot);
        mx = max(mx, te[1]);
        te[tot + 1] = 1e9;
        for(int j = 1; j <= tot; j++) {
            mxn[te[j]] = max(mxn[te[j]], te[j+1]);
        }
    }
    int ans = 1e9;
    for(int i = 0; i <= 100000; i++) {
        ans = min(ans, mx - i);
        mx = max(mx, mxn[i]);
    }
    printf("%d\n", ans);
}

int main() {
    int T = read();
    while(T--) {
        Input();
        Work();
    }
    return 0;
}

E - Qpwoeirut and Vertices

这个是 Kruskal 重构树的模板题，我会单独写一篇来讲 Kruskal 重构树的。