2024.5.15晚训题解

毫无难度啊。

A - Trick Taking
硬模拟就行。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n, t, c[maxn], r[maxn];
signed main(){cin >> n >> t;bool flag = 0;for(int i = 1; i <= n; i++) {cin >> c[i];if(c[i] == t) flag = 1;}int maxx1 = 0, maxx2 = 0, pos1 = -1, pos2 = -1;for(int i = 1; i <= n; i++) {cin >> r[i];if(c[i] == t && maxx1 < r[i]) maxx1 = r[i], pos1 = i;if(c[i] == c[1] && maxx2 < r[i]) maxx2 = r[i], pos2 = i;}cout << (flag ? pos1 : pos2);return 0;
}

B - Same Map in the RPG World
数据范围很小，可以直接枚举(s, t)的所有组合，判断即可。
当然码力弱的人未必能写。

#include<bits/stdc++.h>
using namespace std;
int n, m;
char a[35][35], b[35][35];
bool check(int s, int t) {bool flag = 1;for(int i = 1, j = s; i <= n; i++, j++) {if(j > n) j = 1;for(int k = t, h = 1; h <= m; k++, h++) {if(k > m) k = 1;if(a[i][h] != b[j][k]) {flag = 0;break;}}if(!flag) break;}return flag;
}
signed main(){cin >> n >> m;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)cin >> a[i][j];for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)cin >> b[i][j];for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)if(check(i, j)) {cout << "Yes";return 0;}cout << "No";return 0;
}

C - Cross
直接枚举各个长度的十字架就完事了。
这场看起来相当暴力。

# include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
char c[maxn][maxn];
int n, m, ans[maxn];
int d[][2] = {{-1, -1}, {-1, 1}, {1,  1}, {1,  -1}};
int check(int x, int y) {int ret = 0, t = 1;while(1) {for(int i = 0; i < 4; i++) {int dx = x + d[i][0] * t, dy = y + d[i][1] * t;if(dx < 1 || dx > n || dy < 1 || dy > m) return ret;if(c[dx][dy] == '.') return ret;}ret = t;t++;}
}
signed main() {cin >> n >> m;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)cin >> c[i][j];for(int i = 1; i <= n; i++)for(int j = 0; j < m; j++)if(c[i][j] == '#')ans[check(i, j)]++;for(int i = 1; i <= min(n, m); i++) cout << ans[i] << " ";return 0;
}

D - Cards Query Problem
典型的STL题：
对于操作1和操作2，我们可以开个vector数组，那么把数组 $i$ 放入 $j$ 实际上就可以用v[j].push_back(i)来搞定，输出同理，排个序就完事了。
对于操作3来说有排序和去重操作，所以我们可以用set维护。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n, m;
vector<int> v[maxn];
set<int> s[maxn];
int main() {cin >> n >> m;for(int i = 0; i < m; i++) {int opt, x; cin >> opt >> x;if(opt == 1) {int y; cin >> y;v[y].push_back(x), s[x].insert(y);} else if(opt == 2) {sort(v[x].begin(), v[x].end());for(auto val : v[x]) cout << val << " ";cout << endl;} else {for(auto val : s[x]) cout << val << " ";cout << endl;}}return 0;
}