The 2024 International Collegiate Programming Contest in Hubei Province, China
A - Long Live
题意
思路
当 \(a=1\) 时，\(ab\) 最大
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int x, y;
cin >> x >> y;
cout << 1 << " " << x * y / (__gcd(x, y) * __gcd(x, y)) << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
cin >> __;
while (__--) solve();
return 0;
}
B - Nana Likes Polygons
题意
思路
显然要选\(3\)个点组成三角形，数据很小直接枚举顶点，向量叉积算面积
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int n;
cin >> n;
vector<pii> a(n);
for (auto& [x, y] : a)
{
cin >> x >> y;
}
double ans = 1e18;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (i == j) continue;
for (int k = 0; k < n; k++)
{
if (i == k || j == k) continue;
auto [x1, y1] = a[i];
auto [x2, y2] = a[j];
auto [x3, y3] = a[k];
// 向量
double X1 = 1.0 * x1 - x2;
double Y1 = 1.0 * y1 - y2;
double X2 = 1.0 * x1 - x3;
double Y2 = 1.0 * y1 - y3;
double area = fabs(X1 * Y2 - X2 * Y1) / 2;
if (area >= 1e-6) ans = min(ans, area);
}
}
}
cout << fixed << setprecision(6) << (ans == 1e18 ? -1 : ans) << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
cin >> __;
while (__--) solve();
return 0;
}
E - Spicy or Grilled?
题意
思路
算
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int n, x, a, b;
cin >> n >> x >> a >> b;
cout << x * b + (n - x) * a << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
cin >> __;
while (__--) solve();
return 0;
}
G - Genshin Impact Startup Forbidden II
题意
思路
答奋模拟说是，fzb爆写\(1h^{++}\)
代码
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#include<bits/stdc++.h>
#define int long long
#define PII pair<int, int>
#define endl '\n'
#define inf 1e18
using namespace std;
int dx[4] = { 1, -1, 0, 0 };
int dy[4] = { 0, 0, 1, -1 };
void solve() {
int n;
cin >> n;
vector<vector<int> > G(20, vector<int>(20));
vector<int> f(20 * 20);
for (int i = 1; i <= 19; ++i) {
for (int j = 1; j <= 19; ++j) {
int aim = 19 * (i - 1) + j;
f[aim] = aim;
}
}
function<int(int)> find = [&](int x) {
if (f[x] == x) return x;
return f[x] = find(f[x]);
};
vector<int> sum(20 * 20);
auto merge = [&](int u, int v) {
u = find(u);
v = find(v);
if (u == v) return;
sum[v] += sum[u];
sum[u] = 0;
f[u] = v;
};
vector<vector<int> > sta(20, vector<int>(20));
auto check = [&](int x, int y) {
return (x >= 1 && x <= 19 && y >= 1 && y <= 19);
};
for (int T = 1; T <= n; ++T) {
int x, y;
cin >> x >> y;
set<int> ord;
if (T & 1) sta[x][y] = 1;
else sta[x][y] = 2;
int flag = sta[x][y];
int aim = (x - 1) * 19 + y;
f[aim] = aim;
sum[aim] = 0;
for (int i = 0; i < 4; ++i) {
int cx = x + dx[i];
int cy = y + dy[i];
if (!check(cx, cy)) continue;
if (!sta[cx][cy]) {
sum[aim]++;
}
else {
int u = (cx - 1) * 19 + cy;
u = find(u);
sum[u]--;
}
}
for (int i = 0; i < 4; ++i) {
int cx = x + dx[i];
int cy = y + dy[i];
if (!check(cx, cy)) continue;
if (sta[cx][cy] == sta[x][y]) {
int v = (cx - 1) * 19 + cy;
merge(aim, v);
aim = find(aim);
}
}
for (int i = 0; i < 4; ++i) {
int cx = x + dx[i];
int cy = y + dy[i];
if (!check(cx, cy)) continue;
if (sta[cx][cy] && sta[cx][cy] != sta[x][y]) {
int u = (cx - 1) * 19 + cy;
u = find(u);
if (sum[u] == 0) ord.insert(u);
}
}
int sum1 = 0, sum2 = 0;
for (int xx = 1; xx <= 19; ++xx) {
for (int yy = 1; yy <= 19; ++yy) {
if (!sta[xx][yy]) continue;
int u = (xx - 1) * 19 + yy;
if (ord.count(find(u))) {
for (int k = 0; k < 4; ++k) {
int cx = xx + dx[k];
int cy = yy + dy[k];
if (!check(cx, cy)) continue;
int v = (cx - 1) * 19 + cy;
v = find(v);
if (sta[cx][cy] && sta[cx][cy] == sta[x][y]) {
sum[v]++;
}
}
sta[xx][yy] = 0;
flag == 1 ? sum2++ : sum1++;
f[u] = u;
sum[u] = 0;
}
}
}
aim = find(aim);
if (sum[aim] == 0) {
for (int xx = 1; xx <= 19; ++xx) {
for (int yy = 1; yy <= 19; ++yy) {
if (!sta[xx][yy]) continue;
int u = (xx - 1) * 19 + yy;
if (find(u) == aim) {
for (int k = 0; k < 4; ++k) {
int cx = xx + dx[k];
int cy = yy + dy[k];
if (!check(cx, cy)) continue;
int v = (cx - 1) * 19 + cy;
v = find(v);
if (sta[cx][cy] && sta[cx][cy] != sta[x][y]) {
sum[v]++;
}
}
sta[xx][yy] = 0;
flag == 1 ? sum1++ : sum2++;
f[u] = u;
sum[u] = 0;
}
}
}
}
cout << sum1 << ' ' << sum2 << endl;
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int T = 1;
//cin >> T;
while (T--) {
solve();
}
return 0;
}
H - Genshin Impact Startup Forbidden III
题意
思路
格子中鱼的数量只会是 \(0、1、2、3\)，即把鱼的数量压成四进制
代码
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#include<bits/stdc++.h>
#define int long long
#define PII pair<int, int>
#define endl '\n'
#define inf 1e18
using namespace std;
struct node {
int x, y, cnt;
};
int dx[4] = { 1, -1, 0, 0 };
int dy[4] = { 0, 0, 1, -1 };
void solve() {
int n, m, k;
cin >> n >> m >> k;
vector<int> p4(15);
p4[0] = 1;
for (int i = 1; i <= 10; ++i) {
p4[i] = p4[i - 1] * 4;
}
vector<vector<int> > mp(n + 1, vector<int>(m + 1, 0));
vector<node> t(k + 1);
int now = 0;
for (int i = 1; i <= k; ++i) {
int x, y, cnt;
cin >> x >> y >> cnt;
now += p4[i - 1] * cnt;
t[i] = { x, y, cnt };
mp[x][y] = i;
}
vector<int> f(p4[k], inf);
f[now] = 0;
auto check = [&](int x, int y) {
return (x >= 1 && x <= n && y >= 1 && y <= m);
};
for (int i = now; i; --i) {
if (f[i] == inf) continue;
for (int j = 1; j <= k; ++j) {
if (i / p4[j - 1] % 4) {
auto [x, y, cnt] = t[j];
int st = i;
int p = mp[x][y];
if (p && st / p4[p - 1] % 4) {
st -= p4[p - 1];
}
for (int s = 0; s < 4; ++s) {
int cx = x + dx[s];
int cy = y + dy[s];
if (!check(cx, cy)) continue;
p = mp[cx][cy];
if (p && st / p4[p - 1] % 4) {
st -= p4[p - 1];
}
}
f[st] = min(f[st], f[i] + 1);
for (int s = 0; s < 4; ++s) {
int cx = x + dx[s];
int cy = y + dy[s];
if (!check(cx, cy)) continue;
st = i;
p = mp[cx][cy];
if (p && st / p4[p - 1] % 4) {
st -= p4[p - 1];
}
for (int a = 0; a < 4; ++a) {
int nx = cx + dx[a];
int ny = cy + dy[a];
if (!check(nx, ny)) continue;
p = mp[nx][ny];
if (p && st / p4[p - 1] % 4) {
st -= p4[p - 1];
}
}
f[st] = min(f[st], f[i] + 1);
}
}
}
}
cout << f[0] << endl;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int T = 1;
//cin >> T;
while (T--) {
solve();
}
return 0;
}
J - Points on the Number Axis A
题意
思路
随机取两个点，即每个点被取到的概率相等，最后一个点的期望位置就是 \(\frac 1 n{\sum_{i=1}^n x_i}\)
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
const int mod = 998244353;
int qpow(int base, int a)
{
int res = 1;
while (a)
{
if (a & 1) res = res * base % mod;
base = base * base % mod;
a >>= 1;
}
return res;
}
int inv(int x)
{
return qpow(x, mod - 2) % mod;
}
void solve()
{
int n, sum = 0, x;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> x;
sum = (sum + x) % mod;
}
cout << sum * inv(n) % mod << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--) solve();
return 0;
}
L - LCMs
题意
思路
非互质特判见代码；对于互质的 \(a\) 与 \(b\)，设二者最小质因数为 \(p\) 与 \(q\)，最终的答案就是从 \(a\) 出发， 以 \(2、p、q\) 作为中转点的最短路径
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
const int mod = 998244353;
int f(int x)
{
for (int i = 2; i <= sqrt(x); i++)
{
if (x % i == 0) return i;
}
return x;
}
void solve()
{
int a, b;
cin >> a >> b;
if (a == b) cout << 0 << endl;
else if (b % a == 0) cout << b << endl;
else if (gcd(a, b) != 1) cout << a + b << endl;
else
{
int p = f(a), q = f(b), ans = 1e18;
ans = min(ans, lcm(a, b));
ans = min(ans, lcm(a, p) + lcm(p, b));
ans = min(ans, lcm(a, q) + lcm(q, b));
ans = min(ans, lcm(a, 2) + lcm(2, b));
ans = min(ans, lcm(a, p) + lcm(p, q) + lcm(q, b));
ans = min(ans, lcm(a, p) + lcm(p, 2) + lcm(2, b));
ans = min(ans, lcm(a, 2) + lcm(2, q) + lcm(q, b));
ans = min(ans, lcm(a, p) + lcm(p, 2) + lcm(2, q) + lcm(q, b));
cout << ans << endl;
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
cin >> __;
while (__--) solve();
return 0;
}
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