A. Middle of the Contest

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 200100

int m1,m2,h1,h2;

int main() {
    scanf("%d:%d", &h1, &m1);
    scanf("%d:%d", &h2, &m2);
    int s = h1 * 60 + m1, t = h2 * 60 + m2;
    int mid = (s + t) >> 1, q1 = mid / 60, q2 = mid % 60;
    if(q1 < 10) putchar('0'); printf("%d:",q1);
    if(q2 < 10) putchar('0'); printf("%d",q2);
}

B. Preparation for International Women’s Day

因为\(d_1+d_2=k\)，所以\(d_1+d_2\%k=0\)。
所以模一下存进对应剩余系里面，然后对每个对应的剩余系统计一下即可。

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 200100

int a[N], n, k, vis[N];

int main() {
    memset(vis, 0, sizeof(vis));
    in(n); in(k);
    for(int i = 1; i <= n; ++i) in(a[i]), a[i]%=k, vis[a[i]]++;
    int ans = vis[0] / 2;
    if(k % 2 == 0) ans += vis[k / 2] / 2;
    for(int i = 1; i < (k + 1) / 2; ++i) {
        int j = k - i;
        ans += min(vis[i], vis[j]);
    }
    outn(ans*2);
}

C. Balanced Team

用双指针搞搞就好了（因为有单调性）。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 300010

int n, a[N];

int main() {
    in(n);
    for(int i = 1; i <= n; ++i) in(a[i]);
    sort(a+1,a+n+1); int ans = 0;
    for(int l = 1, r = 1; l <= n; ++l) {
        while(a[r + 1] - a[l] <= 5 && r < n) ++r;
        ans = max(ans, r - l + 1);
    }
    printf("%d\n", ans);
}

D. Zero Quantity Maximization

其实题意就是统计相同的分数有多少个（直接除gcd化成最简分数），然后对于0 0要特殊统计（一定能计入答案）。
注意除0什么的就行了，细节还是挺多的…

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 200100

int a[N], b[N], n, m;
map<pair<int, int>, int>mp;

int main() {
    in(n); int ans = 0, s = 0;
    for(int i = 1; i <= n; ++i) in(a[i]);
    for(int i = 1; i <= n; ++i) {
        int x = read(), g = __gcd(a[i], x);
        if(!x && !a[i]) {
            ++s; continue;
        }
        if(!a[i]) continue;
        ans = max(ans, ++mp[make_pair(a[i]/g, x/g)]);
    }
    printf("%d\n", ans+s);
}

E. K Balanced Teams

设\(f[i][j]\)表示前i个人，选了j个team。
考虑继承\(f[i][j]=f[i-1][j]\)
转移则为\(f[i][j]=max\{f[k-1][j-1]+i-k+1\}(a[i]-a[k]<=5)\)，
注意到k有单调性，那么拿个指针维护一下最远的k就可以优化到\(O(n^2)\)了。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 5100

int vis[N], n, k;
ll a[N], f[N][N];

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) in(a[i]);
    sort(a+1,a+n+1);
    for(int j = 1; j <= k; ++j) {
        for(int l = 1, i = 1; i <= n; ++i) {
            f[i][j] = f[i - 1][j];
            while(l < i && a[i] - a[l] > 5) ++l; 
            f[i][j] = max(f[i][j], f[l - 1][j - 1] + (i - l + 1));
        } 
    }
    printf("%lld\n", f[n][k]);
}
/*
f[i][j]位置i，选了j个team 
*/

F1. Spanning Tree with Maximum Degree

题意杀型题目…题意是让原来度数最大的点，弄出来生成树后，度数还是和原来一样，所以先把与度数最大的那个点连着的边放进生成树里面，然后类似最小生成树那样用并查集判断是否需要这条边，选出n-1条即可。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 200100

int d[N], n, m;
struct edge {
    int u, v;
}e[N];
vector<int>G[N];
int f[N], ans[N][2], cnt;
map<pair<int, int> , bool>mp;

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

int main() {
    in(n), in(m);
    for(int i = 1; i <= m; ++i) {
        in(e[i].u), in(e[i].v); 
        d[e[i].u]++; d[e[i].v]++;
        G[e[i].u].push_back(e[i].v);
        G[e[i].v].push_back(e[i].u);
        if(e[i].u > e[i].v) swap(e[i].u, e[i].v);
    }
    int id = 0, sum = 0;
    for(int i = 1; i <= n; ++i) {
        f[i] = i;
        if(d[i] > sum) sum = d[i], id = i;
    }
    for(int i = 0, len = G[id].size(); i < len; ++i) {
        int x = find(id), y = find(G[id][i]);
        f[y] = x;
        x = id, y = G[id][i];
        if(x > y) swap(x, y);
        ans[++cnt][0] = x, ans[cnt][1] = y;
    }
    for(int i = 1; i <= m; ++i) {
        if(cnt == n - 1) continue;
        if(mp.count(make_pair(e[i].u, e[i].v))) continue;
        int x = find(e[i].u), y = find(e[i].v);
        if(x != y) {
            f[y] = x;
            ans[++cnt][0] = e[i].u;
            ans[cnt][1] = e[i].v;
        }
    }
    for(int i = 1; i <= cnt; ++i) printf("%d %d\n", ans[i][0], ans[i][1]);
    return 0;
}

F2. Spanning Tree with One Fixed Degree

先把除了1以外的节点并起来，那么就会出现几个联通块，显然必须保证1到每个联通块都至少有一条边，那么无解的判定条件就有了，必须有的边不能超过d，1出去的边也不能超过d。然后就按类似最小生成树的流程把剩下的给并起来就好了

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 200100

int d[N], n, m;
struct edge {
    int u, v;
}e[N];
int f[N], b[N][2], D, cnt, tot, ans[N][2], vis[N], t[N];
map<pair<int, int> , bool>mp;

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

int main() {
    in(n), in(m); in(D);
    for(int i = 1; i <= n; ++i) f[i] = i;
    for(int i = 1; i <= m; ++i) {
        in(e[i].u), in(e[i].v); 
        if(e[i].u > e[i].v) swap(e[i].u, e[i].v);
        if(e[i].u != 1 && e[i].v != 1) {
            int x = find(e[i].u), y = find(e[i].v);
            f[y] = x;
        } else b[++tot][0] = e[i].u, b[tot][1] = e[i].v;
    } 
    int now = 0; cnt = 0;
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= tot; ++i) {
        if(!vis[find(b[i][1])]) {
            vis[find(b[i][1])] = 1;
            ans[++now][0] = b[i][0], ans[now][1] = b[i][1];
            mp[make_pair(b[i][0], b[i][1])] = 1;
        }
    }
    if(tot < D) return puts("NO"), 0;
    if(now > D) return puts("NO"), 0;
    int l = 1;
    for(int i = 1; i <= n; ++i) f[i] = i;
    for(int i = 1; i <= now; ++i) {
        f[find(ans[i][1])] = find(ans[i][0]);
    }
    for(int i = now + 1; i <= D; ++i, ++l) {
        while(mp.count(make_pair(b[l][0], b[l][1])) && l < tot) ++l;
        int x = find(b[l][0]), y = find(b[l][1]);
        f[y] = x;
        ans[i][0] = b[l][0], ans[i][1] = b[l][1];
    }
    now = D;
    for(int i = 1; i <= m; ++i) {
        if(now == n - 1) break;
        if(e[i].u == 1 || e[i].v == 1) continue;
        int x = find(e[i].u), y = find(e[i].v);
        if(x != y) {
            f[y] = x;
            ans[++now][0] = e[i].u, ans[now][1] = e[i].v;
        }
    }
    puts("YES");
    for(int i = 1; i < n; ++i) printf("%d %d\n", ans[i][0], ans[i][1]);
    return 0;
}