传送门

A 简单的整除

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace  std;

void solve() {
    int n;
    cin >> n;
    if (n % 2 == 0 || n % 3 == 0 || n % 5 == 0 || n % 7 == 0)cout << "YES" << endl;
    else cout << "NO" << endl;
}


int32_t main() {
    IOS;
    int h_h;
    //cin >> h_h;
    h_h = 1;
    while (h_h--)solve();
    return 0;
}

B 整数划分

就简单的从小到大往后放，当剩余的要小于前一个就加到最后一个一起放

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace  std;

void solve() {
    int n;
    cin >> n;
    for (int i = 1; n; i++) {
        if (n - i <= i && n - i != 0)continue;
        cout << i << ' ';
        n -= i;
    }
    cout << endl;
}


int32_t main() {
    IOS;
    int h_h;
    cin >> h_h;
    //h_h = 1;
    while (h_h--)solve();
    return 0;
}

C 传送阵

就统计数字出现的种类即可

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace  std;

void solve() {
    int n, m;
    cin >> n >> m;
    map<int, int> mp;
    for (int i = 0; i < m * n; i++) {
        int x;
        cin >> x;
        mp[x]++;
    }
    cout << mp.size() << endl;
}


int32_t main() {
    IOS;
    int h_h;
    cin >> h_h;
    //h_h = 1;
    while (h_h--)solve();
    return 0;
}`

D 修改后的和

我们通过题目分析可以知道，一定是要从大到小开始减，二且要先间最右边的，一次往左减，因为减去了这和数会影响这个数后面的结果，总体来说就是一个贪心

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define int long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace  std;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    vector<int> b;
    int sum = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
        if (a[i] >= 0)b.push_back(a[i] * (n - i));
    }
    sort(b.begin(), b.end(), greater<int>());
    //cout << sum << endl;
    //for (auto i : b)cout << i << ' '; cout << endl;
    for (int i = 0; i < min(m, (int)b.size()); i++) {
        sum -= b[i];
    }
    cout << sum << endl;
}


int32_t main() {
    IOS;
    int h_h;
    cin >> h_h;
    //h_h = 1;
    while (h_h--)solve();
    return 0;
}

E 幼稚园的树2

就是简单的推式子，先让树长到超过目标值摈玫瑰裁剪，此后的都是循环，要记得向上取整呢个，而且要超过k，所要去k+1，这里要注意

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace  std;

void solve() {
    int n, m , k, b;
    cin >> n >> m >> k >> b;
    vector<int> a(n + 1), c(n + 1);
    for (int i = 0; i < n; i++)cin >> a[i];
    for (int i = 0; i < n; i++)cin >> c[i];
    //cout << n << ' ' << m << ' ' << k << ' ' << b << endl;
    vector<int> ans;
    m--;
    for (int i = 0; i < n; i++) {
        int x = (k + 1 - a[i] + c[i] - 1) / c[i];
        if (m < x) {
            ans.push_back(a[i] + m * c[i]);
            continue;
        }
        else {
            int xx = (k + 1 - b + c[i] - 1) / c[i];
            int yy = (m - x) % xx;
            ans.push_back(b + c[i]*yy);
        }
    }
    for (auto i : ans)cout << i << ' ';
    cout << endl;
}


int32_t main() {
    IOS;
    int h_h;
    cin >> h_h;
    //h_h = 1;
    while (h_h--)solve();
    return 0;
}

F 最便宜的构建

最大值最小，很经典的二分，我们可以一每条边的全值从小到大排序，一小标进行二分，最后输出权值即可，最关键的就是怎么写check函数，这里使用并查集来判断连通性，也可以使用dfs都行

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se seocnd
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace  std;

struct S{
    int  a, b, c;
    bool operator<(const S &t)const{
        return  c<t.c;
    }
}g[N];
int fa[N];
int n, m, k;
vector<int> gg[N];

int find(int x){
    if(fa[x]!=x)return fa[x]=find(fa[x]);
    return fa[x];
}

void Union(int a, int b) {
    a = find(a);
    b = find(b);
    fa[a] = b;
}

bool check(int x) {
    met_0(fa);
    for (int i = 1; i <= n; i++)fa[i] = i;
    for (int i = 1; i <= x; i++)Union(g[i].a, g[i].b);
    for (int i = 1; i <= k; i++) {
        int node = find(gg[i][0]);
        for (auto j: gg[i]) {
            if (find(j) != node)return false;
        }
    }
    return true;
}

void solve() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
        cin >> g[i].a >> g[i].b >> g[i].c;
    sort(g + 1, g + m + 1);
    cin >> k;
    for (int i = 1; i <= k; i++) {
        int x;
        cin >> x;
        for (int j = 0; j < x; j++) {
            int y;
            cin >> y;
            gg[i].push_back(y);
        }
    }
    int l = 0, r = m - 1;
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid))r = mid;
        else l = mid + 1;
    }
    cout << g[l].c << endl;
}


int32_t main() {
    IOS;
    int h_h;
    //cin >> h_h;
    h_h = 1;
    while (h_h--)solve();
    return 0;
}