每日两题day33

如果你花时间了解我的话，你就会发现，你浪费了一点时间。

每日两题

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一、基础题

题目：P1678 烦恼的高考志愿

思路：

对于每个学生，要求选择一个学校，使得分数差的绝对值最小，并将所有学生的最小分数差累加。可以先将学校分数线排序，然后对每个学生用二分查找找到最接近的学校分数线，计算最小差值并累加。

可能需要的知识：

• 二分：学习网站：OI Wiki，b站: 蛋从几层楼往下丢才会碎？一个视频讲明白【NotOnlySuccess】

代码(c++)：

时间复杂度 O(n \log m)，其中 $n$ 为学生人数，$m$ 为学校数量。

#include <bits/stdc++.h>
using namespace std;int m, n;
vector<int> school, student;void solve() {cin >> m >> n;school.resize(m);student.resize(n);for (int i = 0; i < m; ++i) {cin >> school[i];}for (int i = 0; i < n; ++i) {cin >> student[i];}sort(school.begin(), school.end());long long ans = 0;for (int i = 0; i < n; ++i) {int idx = lower_bound(school.begin(), school.end(), student[i]) - school.begin();int diff = 0;if (idx >= m) {diff = abs(school[m - 1] - student[i]);} else if (idx == 0) {diff = abs(school[0] - student[i]);} else {diff = min(abs(school[idx] - student[i]), abs(school[idx - 1] - student[i]));}ans += diff;}cout << ans << endl;
}int main() {ios::sync_with_stdio(false);cin.tie(nullptr);solve();return 0;
}

二、提高题

题目：P1205 [USACO1.2] 方块转换 Transformations

思路：

纯纯史山大模拟，注意细节即可

代码(c++)：

时间复杂度 O(n^2)

#include <bits/stdc++.h>
using namespace std;int n;
char a[15][15], b[15][15], c[15][15], d[15][15];// 90度顺时针旋转
bool rotate90() {for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)b[j][n - i + 1] = a[i][j];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (b[i][j] != c[i][j]) return false;return true;
}// 180度旋转
bool rotate180() {for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)b[n - i + 1][n - j + 1] = a[i][j];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (b[i][j] != c[i][j]) return false;return true;
}// 270度旋转
bool rotate270() {for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)b[n - j + 1][i] = a[i][j];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (b[i][j] != c[i][j]) return false;return true;
}// 水平翻转
bool reflect() {for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)b[i][n - j + 1] = a[i][j];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (b[i][j] != c[i][j]) return false;return true;
}// 水平翻转后再旋转
bool reflectAndRotate() {reflect();// 复制翻转后的b到afor (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)a[i][j] = b[i][j];if (rotate90()) return true;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)a[i][j] = b[i][j];if (rotate180()) return true;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)a[i][j] = b[i][j];if (rotate270()) return true;return false;
}// 不变
bool noChange() {for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (a[i][j] != c[i][j]) return false;return true;
}void solve() {if (rotate90()) { cout << 1; return; }if (rotate180()) { cout << 2; return; }if (rotate270()) { cout << 3; return; }if (reflect()) { cout << 4; return; }if (reflectAndRotate()) { cout << 5; return; }if (noChange()) { cout << 6; return; }cout << 7;
}int main() {cin >> n;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {cin >> a[i][j];d[i][j] = a[i][j];}for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)cin >> c[i][j];solve();return 0;
}