Codeforces Round #705 (Div. 2)
A - Anti-knapsack
模拟
const int N = 1e5 + 5;
int n, m, _, k, cas;
int v[N];
int main() {
IOS;
for (cin >> _; _; --_) {
cin >> n >> m; VI a;
rep (i, 1, n) v[i] = 1; v[m] = 0;
per (i, n, 1) if (v[i]) {
a.pb(i);
if (i < m) v[m - i] = 0;
}
cout << a.size() << '
';
for (auto &i : a) cout << i << ' '; cout << '
';
}
return 0;
}
B - Planet Lapituletti
模拟
const int N = 1e5 + 5;
int n, m, _, k, cas;
int a[N], c[N];
int main() {
a[0]=1; a[1]=1; a[2]=1;
a[5]=1; a[8]=1;
c[0]=0; c[1]=1; c[2]=5;
c[5]=2; c[8]=8;
for (read(_); _; --_) {
int h, m, hh, mm, nowh, nowm; read(h, m, hh, mm);
bool f = 0;
rep (i, hh, h - 1) {
for(int j = (i == hh ? mm : 0); !f && j < m; ++j) {
if(a[i % 10] && a[i / 10] && a[j % 10] && a[j / 10]) {
nowh = c[j / 10] + c[j % 10] * 10;
nowm = c[i / 10] + c[i % 10] * 10;
if (nowh < h && nowm <m) printf("%02d:%02d
", i, j), f = 1;
}
}
if (f) break;
}
if (!f) printf("00:00
");
}
return 0;
}
C - K-beautiful Strings
n % k != 0 无解
如果 s 本身就是， 直接输出
其他情况，无非是在距离 n 最近的位置 t[i] > s[i], 1 + i <= j <= n 之间插入a 和剩余字符
1 <= j < i, t[i] = s[i]
const int N = 1e5 + 5;
int n, m, _, k, cas;
int a[26];
char s[N];
bool work(int res) {
int c = 0;
rep(i, 0, 25) if (a[i] % k) c += k - a[i] % k;
int cur = n - res - c;
if (cur >= 0 && cur % k == 0) {
rep(i, 1, res) cout << char(s[i]);
rep(i, 1, cur) cout << 'a';
rep(i, 0, 25) if (a[i] % k)
rep(j, 1, k - a[i] % k) cout << (char)(i + 'a');
cout << '
'; return 1;
}
return 0;
}
int main() {
IOS;
for (cin >> _; _; --_) {
cin >> n >> k >> s + 1; memset(a, 0, sizeof a);
if (n % k) { cout << -1 << '
'; continue; }
rep(i, 1, n) ++a[s[i] - 'a']; bool f = 1;
rep(i, 0, 25) if (a[i] % k) f = 0;
if (f) { cout << s + 1 << '
'; continue; }
per(i, n, 1) {
--a[s[i] - 'a']; f = 0;
rep(j, s[i] - 'a' + 1, 25) {
++a[j], s[i] = 'a' + j;
if (work(i)) { f = 1; break; }
--a[j];
}
if (f) break;
}
}
return 0;
}
D - GCD of an Array
人均会动态开点是吧
动态线段树维护 1~n 位置最少有多少个质数 pri[i]
每次操作（最初把每个位置视为 1）最多有7个不同的质数， 复杂度为 (2 * 7 * n * log(n))
map也行, 代码放下面了
const int N = 2e5 + 5, M = 17984 + 5, mod = 1e9 + 7;
struct BIT {
struct node { int l = 0, r = 0, val = 0; } tr[20000000];
int rt[N], tot;
void push_up(int rt) {
tr[rt].val = min(tr[tr[rt].l].val, tr[tr[rt].r].val);
}
void change(int& p, int l, int r, int d, int k) {
if (!p) p = ++tot;
if (l == r) { tr[p].val += k; return; }
int mid = l + r >> 1;
if (d <= mid) change(tr[p].l, l, mid, d, k);
else change(tr[p].r, mid + 1, r, d, k);
push_up(p);
}
int ask(int x) { return tr[x].val; }
} bit;
int n, m, _, k, cas;
int pri[N], tot;
bool v[N];
void init(int n) {
rep(i, 2, n) {
if (!v[i]) pri[++tot] = i;
rep(j, 1, tot) {
if (i > n / pri[j]) break;
v[i * pri[j]] = 1;
if (i % pri[j] == 0) break;
}
}
}
ll qpow(ll a, ll b) {
ll ans = 1;
for (; b; b >>= 1, a = a * a % mod) if (b & 1) ans = ans * a % mod;
return ans;
}
int main() {
IOS; cin >> n >> m; init(2e5); ll ans = 1;
rep(i, 1, n) {
cin >> k;
for (int j = 1; j <= tot && pri[j] <= k / pri[j]; ++j) if (k % pri[j] == 0) {
int cnt = 0;
while (k % pri[j] == 0) k /= pri[j], ++cnt;
bit.change(bit.rt[pri[j]], 1, n, i, cnt);
}
if (k > 1) bit.change(bit.rt[k], 1, n, i, 1);
}
rep(i, 1, tot) ans = ans * qpow(pri[i], bit.ask(bit.rt[pri[i]])) % mod;
rep(i, 1, m) {
int x, k; cin >> x >> k;
for (int j = 1; j <= tot && pri[j] <= k / pri[j]; ++j) if (k % pri[j] == 0) {
int cnt = 0, res = bit.ask(bit.rt[pri[j]]);
while (k % pri[j] == 0) k /= pri[j], ++cnt;
bit.change(bit.rt[pri[j]], 1, n, x, cnt);
ans = ans * qpow(pri[j], bit.ask(bit.rt[pri[j]]) - res) % mod;
}
if (k > 1) {
int res = bit.ask(bit.rt[k]);
bit.change(bit.rt[k], 1, n, x, 1);
ans = ans * qpow(k, bit.ask(bit.rt[k]) - res) % mod;
}
cout << ans << '
';
}
return 0;
}
const int N = 2e5 + 5, mod = 1e9 + 7;
int n, m, _, k, cas;
int pri[N], tot;
multiset<int> st[N];
unordered_map<int, int> h[N];
bool v[N];
void init(int n) {
rep(i, 2, n) {
if (!v[i]) pri[++tot] = i;
rep(j, 1, tot) {
if (i > n / pri[j]) break;
v[i * pri[j]] = 1;
if (i % pri[j] == 0) break;
}
}
}
ll qpow(ll a, ll b) {
ll ans = 1;
for (; b; b >>= 1, a = a * a % mod) if (b & 1) ans = ans * a % mod;
return ans;
}
int main() {
IOS; cin >> n >> m; init(2e5); ll ans = 1;
rep(i, 1, n) {
cin >> k;
for (int j = 1; j <= tot && pri[j] <= k / pri[j]; ++j) if (k % pri[j] == 0) {
int cnt = 0;
while (k % pri[j] == 0) k /= pri[j], ++cnt;
st[pri[j]].insert(cnt); h[i][pri[j]] = cnt;
}
if (k > 1) st[k].insert(1), h[i][k] = 1;
}
rep(i, 1, tot) if (st[pri[i]].size() == n)
ans = ans * qpow(pri[i], *st[pri[i]].begin()) % mod;
rep(i, 1, m) {
int x, k; cin >> x >> k;
for (int j = 1; j <= tot && pri[j] <= k / pri[j]; ++j) if (k % pri[j] == 0) {
int cnt = 0, res = st[pri[j]].size() == n ? *st[pri[j]].begin() : 0;
if (h[x].count(pri[j])) st[pri[j]].erase(st[pri[j]].find(h[x][pri[j]]));
while (k % pri[j] == 0) k /= pri[j], ++cnt;
h[x][pri[j]] += cnt; st[pri[j]].insert(h[x][pri[j]]);
if (st[pri[j]].size() == n) ans = ans * qpow(pri[j], *st[pri[j]].begin() - res) % mod;
}
if (k > 1) {
int res = st[k].size() == n ? *st[k].begin() : 0;
if (h[x].count(k)) st[k].erase(st[k].find(h[x][k]));
h[x][k] += 1; st[k].insert(h[x][k]);
if (st[k].size() == n) ans = ans * qpow(k, *st[k].begin() - res) % mod;
}
cout << ans << '
';
}
return 0;
}