CC BY 4.0 （除特别声明或转载文章外）
如果这篇博客帮助到你，可以请我喝一杯咖啡~
Asteroids
模板题，求两空间多面体重心最近的距离，将重心到所在凸包表面的最短距离相加即可。
蓝书上的原题，详见这里。
Business Center
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
int n, m;
int main()
{
while (~scanf("%d%d", &n, &m))
{
int ans = 1000000001;
for (int i = 1; i <= m; ++i)
{
int x, y;
scanf("%d%d", &x, &y);
int ret = (x * n) % (x + y);
if (ret == 0)
ret = x + y;
ans = min(ans, ret);
}
printf("%d\n", ans);
}
}
Database
#include <bits/stdc++.h>
using namespace std;
int main()
{
for (int n, m, f; cin >> n >> m;)
{
unordered_map<string, set<int>> mp[15];
string s;
getline(cin, s);
for (int i = f = 1; i <= n; ++i)
{
unordered_map<int, int> mmp;
for (int j = 1; j <= m; ++j)
{
getline(cin, s, j < m ? ',' : '\n');
if (!f)
continue;
for (auto it : mp[j][s])
{
if (mmp.count(it))
{
cout << "NO\n"
<< it << ' ' << i << '\n'
<< mmp[it] << ' ' << j << '\n';
f = 0;
break;
}
mmp[it] = j;
}
mp[j][s].insert(i);
}
}
if (f)
cout << "YES\n";
}
}
Funny Language
#include <bits/stdc++.h>
using namespace std;
void cal(const string &s, int c[])
{
fill(c, c + 26, 0);
for (int i = 0; i < s.size(); ++i)
++c[s[i] - 'A'];
}
string s[1023];
int c[1023][31];
int main()
{
for (int n, m; cin >> m >> n;)
{
for (int i = 0; i < n; ++i)
cin >> s[i];
sort(s, s + n);
for (int i = 0; i < n; ++i)
cal(s[i], c[i]);
priority_queue<pair<int, string>> q;
q.push(make_pair(0, ""));
while (m)
{
pair<int, string> pis = q.top();
q.pop();
auto rg = equal_range(s, s + n, pis.second);
if (rg.first == rg.second && !pis.second.empty())
cout << pis.second << '\n', --m;
cal(pis.second, c[n]);
for (pis.second += 'A'; pis.second.back() <= 'Z'; ++pis.second.back())
{
++c[n][pis.second.back() - 'A'];
for (int i = pis.first = 0; i < n; ++i)
{
++pis.first;
for (int j = 0; j < 26; ++j)
if (c[n][j] > c[i][j])
{
--pis.first;
break;
}
}
q.push(pis);
--c[n][pis.second.back() - 'A'];
}
}
}
}
Headshot
#include <bits/stdc++.h>
using namespace std;
int main()
{
for (string s; cin >> s;)
{
int a = s.size(), b = 0, c = 0, d = 0;
for (int i = 0; i < a; ++i)
if (s[i] == '0')
++b, ++c, d += s[(i + 1) % a] == '0';
printf(a * d > b * c ? "SHOOT\n" : a * d < b * c ? "ROTATE\n" : "EQUAL\n");
}
}