More than 2019 cattle off summer school camp (second field) make up part of the problem

A Eddy Walker

Meaning of the questions
have 0 ... n-1, a total of n points in a circle, the starting point is 0, the N and M, said it would go through all the points with a probability of N steps and finally stopped at point M, find this situation.
Idea
hit the table to find the law, the size of a circle of custom calculate the number of times at each stop point, probably infer the probability is 1 / (n-1).
Playing table Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
int vis[maxn], ans[maxn];
int main()
{
    int n, pos, cnt;
    scanf("%d", &n);
    srand((time(0)));
    for(int i = 1; i <= 100000; i++)
    {
        memset(vis, 0, sizeof(vis));
        vis[0] = 1;
        pos = 0;
        cnt = 1;
        while(cnt < n)
        {
            int moving = rand() % 2 ? 1 : -1;
            pos = (pos + moving + n) % n;
            if(!vis[pos])
            {
                vis[pos] = 1;
                cnt++;
            }
            if(cnt == n)
            {
                ans[pos]++;
            }
        }
    }
    for(int i = 0;i < n; i++)
        printf("%d ", ans[i]);
    printf("\n");
    return 0;
}

AC Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
ll qpow(ll a, ll b)
{
    ll ans = 1;
    a = a % mod;
    while (b) {
        if(b&1) ans = ans * a% mod;
        a = a * a % mod;
        b = b >> 1;
    }
    return ans;
}
int main()
{
    int t;
    cin >> t;
    ll ans = 1;
    while (t--) {
        ll n, m;
        cin >> n >> m;
        if(n == 1 && m == 0) {
            printf("%lld\n", ans);
            continue;
        }
        else if(m == 0) {
            ans = 0;
            printf("%lld\n", ans);
            continue;
        }
        printf("%lld\n", ans = ans * qpow(n - 1, mod - 2) % mod);
    }
    return 0;
}

F Partition problem

Meaning of the questions
2N individuals, divided into two teams, seeking the maximum total competitive value.
Ideas
violence packet, but will time out, it still needs pretreatment.

#include <bits/stdc++.h>
using namespace std;
const int maxn = 30 +7;
typedef long long ll;
int m, w[maxn][maxn];
ll sum[maxn][maxn], ans = 0;
int A;
void dfs(int x, int xx, ll v)//x表示选到第x个，xx表示选了xx个
{
    if(xx == m) {
        for (int i = 1; i < x; i++) {
            if(A &(1<<i))
                v += sum[i][2 * m] - sum[i][x - 1];
        }
        if(v > ans) ans = v;
        return;
    }
    if(x - xx - 1 == m) {
        for (int i = 1; i < x; i++) {
            if(!(A&(1<<i)))
                v += sum[i][2 * m] - sum[i][x - 1];
        }
        if(v > ans)
            ans = v;
        return;
    }
    //不选
    ll t = 0;
    for (int i = 1; i < x; i++) {
        if(A&(1<<i)) t += w[i][x];
    }
    dfs(x + 1, xx, v + t);
    //选
    t = 0;
    for (int i = 1; i < x; i++) {
        if(!(A&(1<<i)))
            t += w[i][x];
    }
    A|= 1<<x;
    dfs(x + 1, xx + 1, v + t);
    A &= ~(1<<x);
}
int main()
{
    scanf("%d", &m);
    for (int i = 1; i <= 2 * m; i++) {
        for (int j = 1; j <= 2 * m; j++) {
            scanf("%d", &w[i][j]);
        }
    }
    for (int i = 1; i <= 2 * m; i++) {
        for (int j = 1; j <= 2 * m; j++) {
            sum[i][j] = sum[i][j - 1] + w[i][j];//预处理
        }
    }

    A |= 1<<1;
    dfs(2, 1, 0LL);
    cout << ans << endl;
}

H Second Large Rectangle

Meaning of the questions
are not demanding strict second large rectangle.
Thinking
suspension wires fajitas monotonous stack.
Learn hanging line method
paper: Talking to solve the biggest problem with the rectangular sub-maximizing ideas

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 +  7;
typedef long long ll;//g是原图 g[i][j] = 1表示所求的最大子矩形的组成元素
int g[maxn][maxn], h[maxn][maxn];
priority_queue<int, vector<int>, greater<int> > q;
void add(int x)
{
    q.push(x);
    while (q.size() > 2) q.pop();
}
int st[maxn];
int width[maxn];
int main()
{
    int n, m;
    int ans = 0;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%1d", &g[i][j]);
            h[i][j] = g[i][j] ? h[i - 1][j] + 1: 0;
        }
        h[i][m + 1] = 0;
    }
    q.push(0); q.push(0);
    for (int i = 1; i <= n; i++) {
        int top = 0;
        for (int j = 1; j <= m + 1; j++) {
            if(st[top] < h[i][j]) {
                st[++top] = h[i][j];
                width[top] = 1;
            }
            else {
                int w = 0;
                while (st[top] > h[i][j]) {
                    w += width[top];
                    add(st[top] * w);
                    add(st[top] * (w - 1));
                    add((st[top] - 1) * w);
                    top--;
                }
                st[++top] = h[i][j];
                width[top] = w + 1;
            }
        }
    }
    printf("%d\n", q.top());
    return 0;
}