A - STring
用一个栈模拟即可。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') ch = getchar(), a = -1;
else a = 1;
x = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + ch - '0';
return a * x;
}
const int N = 2e5 + 10;
char str[N], stk[N];
int n, tp;
int main()
{
scanf("%s", str + 1);
n = strlen(str + 1);
for (int i = 1; i <= n; ++i)
{
if (tp > 0 && stk[tp] == 'S' && str[i] == 'T') --tp;
else stk[++tp] = str[i];
}
printf("%d
", tp);
return 0;
}
B - Minimum Sum
分治一下，计算一个数作为最小值的贡献。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') ch = getchar(), a = -1;
else a = 1;
x = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + ch - '0';
return a * x;
}
const int N = 2e5 + 10;
int a[N];
long long ans;
void solve(int l, int r)
{
if (l == r)
{
ans += a[l];
return;
}
long long res = 0;
int mid = (l + r) >> 1;
vector<int> L, R;
for (int i = mid; i >= l; --i)
{
if (!SZ(L) || a[L[SZ(L) - 1]] > a[i])
L.pb(i);
}
for (int i = mid + 1; i <= r; ++i)
{
if (!SZ(R) || a[R[SZ(R) - 1]] > a[i])
R.pb(i);
}
L.pb(l - 1);
R.pb(r + 1);
for (int i = 0, j = 0; i < SZ(L) - 1; ++i)
{
while (a[R[j]] > a[L[i]] && j < SZ(R) - 1) ++j;
res += 1LL * (R[j] - mid - 1) * (L[i] - L[i + 1]) * a[L[i]];
}
for (int i = 0, j = 0; i < SZ(R) - 1; ++i)
{
while (a[R[i]] < a[L[j]] && j < SZ(L) - 1) ++j;
res += 1LL * (mid - L[j]) * (R[i + 1] - R[i]) * a[R[i]];
}
ans += res;
solve(l, mid);
solve(mid + 1, r);
}
int n;
int main()
{
gi(n);
for (int i = 1; i <= n; ++i) gi(a[i]);
solve(1, n);
printf("%lld
", ans);
return 0;
}
C - Tree Restoring
先把直径找出来，其它的挂在直径上就好了。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') ch = getchar(), a = -1;
else a = 1;
x = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + ch - '0';
return a * x;
}
int n, a[110];
int main()
{
gi(n);
for (int i = n; i; --i) gi(a[i]);
sort(a + 1, a + n + 1);
reverse(a + 1, a + n + 1);
int len = a[1];
int mn = len / 2 + 1;
for (int i = len; i > len / 2; --i)
{
int cnt = 0;
for (int j = 1; j <= n; ++j)
{
if (a[j] == i) ++cnt, a[j] = 0;
if (cnt == 2) break;
}
if (cnt < 2)
{
puts("Impossible");
return 0;
}
}
if (!(len & 1))
{
mn = len / 2;
int cnt = 0;
for (int j = 1; j <= n; ++j)
{
if (a[j] == (len >> 1)) ++cnt, a[j] = 0;
if (cnt == 1) break;
}
if (cnt < 1)
{
puts("Impossible");
return 0;
}
}
for (int i = 1; i <= n; ++i)
if (a[i] && a[i] <= mn)
{
puts("Impossible");
return 0;
}
puts("Possible");
return 0;
}
D - ~K Perm Counting
我们发现可以放置的可以画出一个二分图，然后容斥一下就好了。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') ch = getchar(), a = -1;
else a = 1;
x = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + ch - '0';
return a * x;
}
const int mod = 924844033;
int f[2010][2010][2], g[2010], t[2010], ans;
int main()
{
int n, k;
gii(n, k);
f[0][0][0] = 1;
for (int i = 1; i <= n; ++i)
{
for (int j = 0; j <= n; ++j)
{
f[i][j][0] = (f[i][j][0] + f[i - 1][j][0]) % mod;
f[i][j][0] = (f[i][j][0] + f[i - 1][j][1]) % mod;
if (j) f[i][j][1] = f[i - 1][j - 1][0];
}
}
g[0] = 1;
for (int i = 1; i <= k; ++i)
{
int len = (n - i) / k;
for (int i = 0; i <= n; ++i)
t[i] = g[i], g[i] = 0;
for (int j = 0; j <= len; ++j)
for (int l = 0; l <= n - j; ++l)
g[j + l] = (g[j + l] + 1LL * t[l] * (f[len][j][0] + f[len][j][1])) % mod;
for (int i = 0; i <= n; ++i)
t[i] = g[i], g[i] = 0;
for (int j = 0; j <= len; ++j)
for (int l = 0; l <= n - j; ++l)
g[j + l] = (g[j + l] + 1LL * t[l] * (f[len][j][0] + f[len][j][1])) % mod;
}
for (int i = n, fac = 1, j = (n & 1) ^ 1; ~i; --i, fac = 1LL * fac * (n - i) % mod, j ^= 1)
{
if (j) ans = (ans + 1LL * g[i] * fac) % mod;
else ans = (ans - 1LL * g[i] * fac) % mod;
}
if (ans < 0) ans += mod;
printf("%d
", ans);
return 0;
}
E - Sugigma: The Showdown
如果有一条A中的边在B中长度大于2，只要A走到肯定赢，A能走到的点就是A要在B之前走到，算一下就好了。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') ch = getchar(), a = -1;
else a = 1;
x = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + ch - '0';
return a * x;
}
const int N = 2e5 + 10;
int n;
VI A[N], B[N];
int fa[N], in[N], out[N], dep[N], dfs_clock;
void dfsB(int u)
{
in[u] = ++dfs_clock;
for (auto v : B[u])
{
if (v == fa[u]) continue;
fa[v] = u;
dep[v] = dep[u] + 1;
dfsB(v);
}
out[u] = dfs_clock;
}
bool check(int u, int v)
{
if (in[u] <= in[v] && in[v] <= out[u])
{
if (fa[v] == u) return 1;
if (fa[fa[v]] == u) return 1;
return 0;
}
if (in[v] <= in[u] && in[u] <= out[v])
{
if (fa[u] == v) return 1;
if (fa[fa[u]] == v) return 1;
return 0;
}
if (fa[u] == fa[v]) return 1;
return 0;
}
bool vis[N];
int d[N];
void bfsA(int s)
{
static int q[N]; int l = 0, r = 0;
q[r++] = s; vis[s] = 1;
while (l < r)
{
int u = q[l++];
for (auto v : A[u])
{
if (vis[v]) continue;
d[v] = d[u] + 1;
if (d[v] < dep[v] && !vis[v])
{
vis[v] = 1;
q[r++] = v;
}
}
}
}
int rootA, rootB;
bool win[N];
int main()
{
giii(n, rootA, rootB);
for (int i = 1; i < n; ++i)
{
int u, v;
gii(u, v);
A[u].pb(v);
A[v].pb(u);
}
for (int i = 1; i < n; ++i)
{
int u, v;
gii(u, v);
B[u].pb(v);
B[v].pb(u);
}
dfsB(rootB);
for (int i = 1; i <= n; ++i)
{
for (auto j : A[i])
if (!check(i, j)) win[i] = win[j] = 1;
}
bfsA(rootA);
int ans = 0;
for (int i = 1; i <= n; ++i)
{
if (vis[i] && win[i])
{
puts("-1");
return 0;
}
if (vis[i]) ans = max(ans, dep[i] * 2);
}
printf("%d
", ans);
return 0;
}
F - Many Easy Problems
分开算每个点贡献，可以用容斥做，最后发现这个式子是一个卷积。NTT一下就好了。
//waz
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
int F()
{
char ch;
int x, a;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
if (ch == '-') ch = getchar(), a = -1;
else a = 1;
x = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
x = (x << 1) + (x << 3) + ch - '0';
return a * x;
}
const int N = 1 << 20;
const int mod = 924844033;
const int g = 5;
int fpow(int a, int x)
{
int ret = 1;
for (; x; x >>= 1)
{
if (x & 1) ret = 1LL * ret * a % mod;
a = 1LL * a * a % mod;
}
return ret;
}
int n;
vector<int> edge[N];
int cnt[N], siz[N], fac[N], ifac[N];
int C(int n, int m)
{
if (n < m || m < 0) return 0;
return 1LL * fac[n] * ifac[n - m] % mod * ifac[m] % mod;
}
void dfs(int u, int fa)
{
siz[u] = 1;
for (auto v : edge[u])
{
if (v == fa) continue;
dfs(v, u);
siz[u] += siz[v];
}
for (auto v : edge[u])
{
if (v == fa)
{
++cnt[n - siz[u]];
continue;
}
++cnt[siz[v]];
}
}
void dft(int *a, int n, int sig)
{
for (int i = 0, j = 0; i < n; ++i)
{
if (i > j) swap(a[i], a[j]);
for (int l = n >> 1; (j ^= l) < l; l >>= 1);
}
for (int i = 1; i < n; i <<= 1)
{
int m = i << 1;
int w = fpow(g, (mod - 1) / m);
if (sig == -1) w = fpow(w, mod - 2);
for (int j = 0; j < n; j += m)
{
int v = 1;
for (int k = j; k < j + i; ++k, v = 1LL * v * w % mod)
{
int x = a[k], y = 1LL * a[k + i] * v % mod;
a[k] = (x + y) % mod;
a[k + i] = (x - y + mod) % mod;
}
}
}
if (sig == -1)
{
int invn = fpow(n, mod - 2);
for (int i = 0; i < n; ++i) a[i] = 1LL * a[i] * invn % mod;
}
}
int main()
{
gi(n);
for (int i = 1; i < n; ++i)
{
int u, v;
gii(u, v);
edge[u].pb(v);
edge[v].pb(u);
}
dfs(1, 0);
fac[0] = 1;
for (int i = 1; i <= n; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
ifac[n] = fpow(fac[n], mod - 2);
for (int i = n; i; --i) ifac[i - 1] = 1LL * ifac[i] * i % mod;
static int f[N], g[N];
for (int i = 0; i <= n; ++i) g[i] = ifac[n - i];
for (int i = 0; i <= n; ++i) f[i] = 1LL * cnt[i] * fac[i] % mod;
int m = 1;
for (; m <= (n << 1); m <<= 1);
dft(f, m, 1), dft(g, m, 1);
for (int i = 0; i < m; ++i) f[i] = 1LL * f[i] * g[i] % mod;
dft(f, m, -1);
for (int k = 1; k <= n; ++k)
{
int ans = n;
ans = 1LL * ans * C(n, k) % mod;
ans = (ans - 1LL * ifac[k] * f[n + k]) % mod;
if (ans < 0) ans += mod;
printf("%d
", ans);
}
return 0;
}