Codeforces Round #641 (Div. 1) 题解
Visits: 459
Orac and LCM
每个质因子考虑一下即可。
namespace shai {
const int N = 1e6 + 7;
int p[N], v[N], phi[N], miu[N];
inline void init(int n) {
v[1] = phi[1] = miu[1] = 1;
for (int i = 2; i <= n; i++) {
if (!v[i]) p[++p[0]] = v[i] = i, phi[i] = i - 1, miu[i] = -1;
for (int j = 1; j <= p[0] && i * p[j] <= n && p[j] <= v[i]; j++)
v[i*p[j]] = p[j],
phi[i*p[j]] = phi[i] * (p[j] - 1 + (p[j] == v[i])),
miu[i*p[j]] = p[j] == v[i] ? 0 : -miu[i];
}
}
}
using shai::p;
const int N = 1e5 + 7, M = 2e5 + 7;
int n, a[N], c[M];
pq<int> q[M];
inline vector<pair<ll, int>> divide(ll n) {
vector<pair<ll, int>> p;
for (ll i = 2; i * i <= n; i++)
if (!(n % i)) {
p.pb(mp(i, 0));
while (!(n % i)) ++p.back().se, n /= i;
}
if (n > 1) p.pb(mp(n, 1));
return p;
}
int main() {
rd(n), rda(a, n);
shai::init(2e5);
for (int i = 1; i <= n; i++) {
auto p = divide(a[i]);
for (auto o : p) {
q[o.fi].push(o.se);
if (q[o.fi].size() > 2u) q[o.fi].pop();
++c[o.fi];
}
}
ll ans = 1;
for (int i = 1; i <= p[0]; i++) {
int x = p[i];
if (c[x] == n) {
int y = q[x].top();
while (y--) ans *= x;
}
if (c[x] == n - 1) {
if (q[x].size() > 1u) q[x].pop();
int y = q[x].top();
while (y--) ans *= x;
}
}
print(ans);
return 0;
}
Orac and Medians
首先判一些特殊情况。
一般情况就是判是否存在连续三个数中有两个 $\ge k$。
const int N = 1e5 + 7;
int n, k, a[N];
inline void solve() {
rd(n, k), rda(a, n);
vi p;
for (int i = 1; i <= n; i++)
if (a[i] == k) p.pb(i);
if (!p.size()) return prints("no");
if (n == 1) return prints("yes");
if (n == 2) return prints(a[1] >= k && a[2] >= k ? "yes" : "no");
for (int i = 1; i <= n - 2; i++) {
int t = (a[i] >= k) + (a[i+1] >= k) + (a[i+2] >= k);
if (t >= 2) return prints("yes");
}
prints("no");
}
int main() {
int T;
rd(T);
while (T--) solve();
return 0;
}
Orac and Game of Life
随便 BFS 一下。
const int N = 1e3 + 7;
ll inf = 1e18 + 1;
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
int n, m, t;
ll d[N][N];
char s[N][N];
int main() {
rd(n, m, t);
for (int i = 1; i <= n; i++) rds(s[i], m);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
d[i][j] = inf;
queue<pi> q;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
bool ok = 0;
for (int o = 0; o < 4; o++) {
int x = i + dx[o], y = j + dy[o];
if (x < 1 || x > n || y < 1 || y > m) continue;
if (s[x][y] == s[i][j]) ok = 1;
}
if (ok) d[i][j] = 0, q.push(mp(i, j));
}
while (q.size()) {
int i = q.front().fi, j = q.front().se;
q.pop();
for (int o = 0; o < 4; o++) {
int x = i + dx[o], y = j + dy[o];
if (x < 1 || x > n || y < 1 || y > m) continue;
if (d[x][y] > d[i][j] + 1) d[x][y] = d[i][j] + 1, q.push(mp(x, y));
}
}
while (t--) {
int x, y;
ll z;
rd(x, y);
rd(z);
if (z <= d[x][y]) print(s[x][y] - '0');
else {
int k = s[x][y] - '0';
z -= d[x][y];
k ^= z & 1;
print(k);
}
}
return 0;
}
Slime and Biscuits
题目太神仙了,先咕咕咕着。
Slime and Hats
题目太神仙了,先咕咕咕着。
Slime and Sequences
题目太神仙了,先咕咕咕着。