T1：Leyland Number
模拟
代码实现
a, b = map(int, input().split())
print(a**b+b**a)
T2：Longest Palindrome
模拟
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
bool isPalindrome(string s) {
string t = s;
reverse(t.begin(), t.end());
return s == t;
}
int main() {
string s;
cin >> s;
int n = s.size();
int ans = 0;
rep(r, n)rep(l, r+1) {
string ns = s.substr(l, r-l+1);
if (isPalindrome(ns)) ans = max(ans, r-l+1);
}
cout << ans << '\n';
return 0;
}
T3：Slot Strategy 2 (Easy)
暴搜这 \(3\) 个转盘停止旋转的时间
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
int m;
cin >> m;
vector<string> s(3);
rep(i, 3) cin >> s[i];
const int INF = 1001001001;
int ans = INF;
rep(t0, 300)rep(t1, 300)rep(t2, 300) {
if (t0 == t1) continue;
if (t0 == t2) continue;
if (t1 == t2) continue;
if (s[0][t0%m] != s[1][t1%m]) continue;
if (s[0][t0%m] != s[2][t2%m]) continue;
ans = min(ans, max({t0, t1, t2}));
}
if (ans == INF) ans = -1;
cout << ans << '\n';
return 0;
}
或者暴搜这 \(3\) 个转盘停止旋转的顺序以及以哪个数停止
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
int m;
cin >> m;
vector<string> s(3);
rep(i, 3) cin >> s[i];
const int INF = 1001001001;
int ans = INF;
vector<int> p = {0, 1, 2};
rep(d, 10) {
char c = '0'+d;
do {
int t = -1;
rep(i, 3) {
t++;
while (t < 300 and s[p[i]][t%m] != c) t++;
}
if (t < 300) ans = min(ans, t);
} while (next_permutation(p.begin(), p.end()));
}
if (ans == INF) ans = -1;
cout << ans << '\n';
return 0;
}
T4：Relative Position
显然我们只需得知点 \(1\) 所在的连通分量中的点的位置
注意原图可能有环，所以不能跑拓扑排序
这里我们可以用 \(\operatorname{dfs}\) 或 \(\operatorname{bfs}\) 解决
注意需要存无向边，因为从点 \(1\) 出发不一定能走到属于同一个连通分量中的某个点
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
struct Edge {
int to, x, y;
Edge(int to=-1, int x=0, int y=0): to(to), x(x), y(y) {}
};
int main() {
int n, m;
cin >> n >> m;
vector<vector<Edge>> g(n);
rep(i, m) {
int a, b, x, y;
cin >> a >> b >> x >> y;
--a; --b;
g[a].emplace_back(b, x, y);
g[b].emplace_back(a, -x, -y);
}
const ll INF = 1e18;
vector<ll> x(n, INF), y(n, INF);
x[0] = y[0] = 0;
queue<int> q;
q.push(0);
while (q.size()) {
int v = q.front(); q.pop();
for (auto [u, dx, dy] : g[v]) {
if (x[u] != INF) continue;
x[u] = x[v]+dx;
y[u] = y[v]+dy;
q.push(u);
}
}
rep(i, n) {
if (x[i] == INF) puts("undecidable");
else cout << x[i] << ' ' << y[i] << '\n';
}
return 0;
}