DL24 10.6

$T1$：$20$分暴力。直接计算每个点左右所能到达的最大值，因为有些点可能在之后会被填平，所以左右每个点都需要记录并排序。然后$dfs$枚举所有选择填平$k$个点的方案，计算即可。$O(max(C(20,i))n$²$)$

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
const int mod = 1e9 + 7;
int n, k, h[30], vis[30], q[30];
ll ans;
struct Max {int maxl[30], maxr[30], maxln[30], maxrn[30];}stu[30];
struct T {int x, num;}t[30];
int cmp(T a, T b) {return a.x > b.x;}
void solve()
{
    ll sum = 0;
    for(int i = 1; i <= n; ++i) {h[i] = q[i]; if(vis[i]) h[i] = 0;}
    for(int i = 1; i <= n; ++i)
    {
        int Maxl = 0, Maxr = 0;
        for(int j = 1; j < i; ++j) if(!vis[stu[i].maxln[j]]) {Maxl = stu[i].maxl[j]; break;};
        for(int j = 1; j <= n - i; ++j) if(!vis[stu[i].maxrn[j]]) {Maxr = stu[i].maxr[j]; break;}
        if(h[i] < Maxl && h[i] < Maxr) sum += (min(Maxl, Maxr) - h[i]);
    }
    if(sum % 2 == 0) ++ans;
    return;
}
void dfs(int x, int num)
{
    if(n - num < k - x) return;
    if(x > k) {solve(); return;}
    for(int i = num; i <= n; ++i)
    {
        if(!vis[i])
        {
            vis[i] = 1;
            dfs(x + 1, i + 1);
            vis[i] = 0;
        }
    }
    return;
}
int main()
{
    freopen("rain.in", "r", stdin), freopen("rain.out", "w", stdout);
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= n; ++i) scanf("%d", &h[i]), q[i] = h[i];
    for(int i = 1; i <= n; ++i)
    {
        int z = 0;
        for(int j = 1; j < i; ++j) t[++z].x = h[j], t[z].num = j;
        sort(t + 1, t + z + 1, cmp);
        for(int j = 1; j <= z; ++j) stu[i].maxl[j] = t[j].x, stu[i].maxln[j] = t[j].num;
        z = 0;
        for(int j = n; j > i; --j) t[++z].x = h[j], t[z].num = j;
        sort(t + 1, t + z + 1, cmp);
        for(int j = 1; j <= z; ++j) stu[i].maxr[j] = t[j].x, stu[i].maxrn[j] = t[j].num;
    }
    dfs(1, 1);
    printf("%lld", ans % mod);
    return 0;
}

$T3$：$40$分暴力。对于每一行的每一块土地，挖掘机可以用钻头进行覆盖，贪心每次覆盖最左边的土地，计算即可。$O(hw$²$)$

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int h, w, k, q;
char s[20][100010];
int solve(int d, int l, int r)
{
    int ans = 0;
    for(int i = 1; i <= d; ++i)
        for(int j = l; j <= r; ++j)
            if(s[i][j] == 'X') 
                ++ans, j = j + k - 1;
    return ans;
}
int main()
{
    freopen("blueshit.in", "r", stdin), freopen("blueshit.out", "w", stdout);
    scanf("%d %d %d %d", &h, &w, &k, &q);
    for(int i = 1; i <= h; ++i) scanf("%s", s[i] + 1);
    for(int i = 1, d, l, r; i <= q; ++i) scanf("%d %d %d", &d, &l, &r), printf("%d\n", solve(d, l, r));
    return 0;
}

$T4$：$20$分暴力，枚举$k$，再枚举$A$和$B$的每一个子串，通过记录$f[k][i][j]$表示长度为$k$且$A$中以$i$开头，$B$中以$j$开头谁获胜进行转移，预处理$f[0][i][j]$。$O(k|A||B|)$

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int lx, ly, f[110][110][110], wx, wy, wp, all;
char x[110], y[110];
int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
int main()
{
    freopen("game.in", "r", stdin), freopen("game.out", "w", stdout);
    scanf("%s %s", x + 1, y + 1);
    lx = strlen(x + 1), ly = strlen(y + 1);
    for(int i = 1; i <= lx; ++i)
    {
        for(int j = 1; j <= ly; ++j)
        {
            if(x[i] < y[j]) f[0][i][j] = 1, ++wx;
            else if(x[i] > y[j]) f[0][i][j] = -1, ++wy;
            else f[0][i][j] = 0, ++wp;
        }
    }
    all = wx + wy + wp;
    printf("%d/%d %d/%d %d/%d\n", wx / gcd(wx, all), all / gcd(wx, all), wp / gcd(wp, all), all / gcd(wp, all), wy / gcd(wy, all), all / gcd(wy, all));
    for(int k = 1; k < min(lx, ly); ++k)
    {
        wx = 0, wy = 0, wp = 0;
        for(int i = 1, j = i + k; i <= lx && j <= lx; ++i, ++j)
        {
            for(int s = 1, t = s + k; s <= ly && t <= ly; ++s, ++t)
            {
                if(f[k - 1][i][s] == 1) ++wx, f[k][i][s] = 1;
                else if(f[k - 1][i][s] == -1) ++wy, f[k][i][s] = -1;
                else
                {
                    if(x[j] < y[t]) ++wx, f[k][i][s] = 1;
                    else if(x[j] > y[t]) ++wy, f[k][i][s] = -1;
                    else ++wp, f[k][i][s] = 0;
                }
            }
        }
        all = wx + wy + wp;
        printf("%d/%d %d/%d %d/%d\n", wx / gcd(wx, all), all / gcd(wx, all), wp / gcd(wp, all), all / gcd(wp, all), wy / gcd(wy, all), all / gcd(wy, all));
    }
    return 0;
}