2022NOIP A层联测32
四处行走 (walk)
比较板的题，直接上\(EXCRT\)，然而我考场并不会
code
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0; char f = 1; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 0;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (!f) x = -x;
}
int stk[40], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
const int maxn = 2e5 + 10;
#define int ll
int n, m;
ll l, r, lcm;
int c[maxn];
vector <pair<int, int> > a[maxn];
ll gcd (ll x, ll y)
{
while (y)
{
int t = x; x = y;
y = t % x;
}
return x;
}
ll LCM (ll x, ll y)
{
return x * y / gcd (x, y);
}
int exgcd (int A, int B, int &x, int &y)
{
if (!B)
{
x = 1, y = 0;
return A;
}
int d = exgcd (B, A % B, y, x);
y -= A / B * x;
return d;
}
int EXCRT (int id)
{
int ans = 0, M = 1;
for (auto i : a[id])
{
int p;
int x, y, d = exgcd (M, i.second, x, y);
if ((i.first - ans) % d) return -1;
int b = i.second / d, c = (i.first - ans) / d;
p = (x * c % b + b) % b;
ans += p * M;
M = LCM (M, i.second);
ans = (ans % M + M) % M;
}
return ans;
}
mizuki main ()
{
freopen ("walk.in", "r", stdin);
freopen ("walk.out", "w", stdout);
in (n), in (m);
in (l), in (r);
lcm = 1;
for (int i = 1, x; i <= n; ++ i)
{
in (x);
lcm = LCM (x, lcm);
for (int J = 1, y; J <= x; ++ J) in (y), a[y].push_back ({J, x});
}
memset (c, -1, sizeof (c));
for (int i = 1; i <= m; ++ i) if ((int) a[i].size () == n) c[i] = EXCRT (i);
ll ans = 0;
for (int i = 1; i <= m; ++ i)
{
if (c[i] != -1)
{
int L = max (ceil (1.0 * (l - c[i]) / lcm), 0.0), R = max ((r - c[i]) / lcm, 0ll);
ans += (R - L + 1);
}
}
ot (ans, 1);
return love;
}
鸟之诗 (air)
比较J的题，它已经给分解好了，而质因数相互独立没有什么影响，于是就可以直接计算每个数的当前质因数的次数应该填什么，发现所有数的当前质因子的次数的最小值必须是它给出的\(a_i\)，最大值必须是\(b_i\)，于是就可以进行一个小容斥
先用全部的情况\(qpow(b_i - a_i +1, n)\)减去一个\(a_i\)也没有的情况\(qpow(b_i - a_i,n)\)减去一个\(b_i\)也没有的情况\(qpow(b_i - a_i,n)\)，发现两个都不选的情况多减了一次，再加上\(qpow (b_i - a_i - 1, n)\)就彳亍
code
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0; char f = 1; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 0;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (!f) x = -x;
}
int stk[40], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
const int maxn = 1e6 + 10;
const int mod = 998244353;
#define int ll
int n, m;
int a[maxn], b[maxn];
ll qwq (ll x, ll y)
{
ll res = 1;
while (y)
{
if (y & 1) res = res * x % mod;
x = x * x % mod;
y >>= 1;
}
return res;
}
mizuki main ()
{
in (m), in (n);
n %= (mod - 1);
for (int i = 1; i <= m; ++ i) in (a[i]);
for (int J = 1; J <= m; ++ J) in (b[J]);
ll ans = 1;
for (int i = 1; i <= m; ++ i)
{
if (b[i] == a[i]) continue;
ll k = ((qwq (b[i] - a[i] + 1, n) - qwq (b[i] - a[i], n) * 2ll % mod + qwq (b[i] - a[i] - 1, n)) % mod + mod) % mod;
ans = (ans * k) % mod;
}
ot (ans);
return love;
}
核心共振 (dimension)
非常J的题，通过\(\text{pdf}\)（或者像\(lyin\)一样巨）就可以知道他的递推式是 \(f(i, J) = f(i - 1, J) + f(i - 1,J - 1)\)，发现和组合数一样，于是就可以当做组合数做了，然而要求的是\(\sum\limits_{i=0}^{n}f(i,m)\)，于是手动打了个表发现这玩意等于\(f(n+1, m+1)-1\)
考虑求\(f(n, m)\)
第3步转化是因为在循环的过程中\(\large\binom{n\mod p}{i\mod p}\)会完整地重复计算\(\left \lfloor \frac{m}{p} \right \rfloor - 1\)次，所以直接提出来枚举\(\left \lfloor \frac{i}{p} \right\rfloor\)，然后加上多出来的不完整的部分，然后就可以递归计算了
code
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0; char f = 1; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 0;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (!f) x = -x;
}
int stk[40], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
const int maxn = 2e7 + 10;
#define int ll
int n, m, mod;
ll qwq (ll x, ll y)
{
ll res = 1;
while (y)
{
if (y & 1) res = res * x % mod;
x = x * x % mod;
y >>= 1;
}
return res;
}
int p[maxn], ny[maxn];
ll C (int x, int y)
{
if (!y) return 1;
if (y > x) return 0;
return p[x] * ny[y] % mod * ny[x - y] % mod;
}
ll luc (int x, int y)
{
if (!x || !y) return 1;
return luc (x / mod, y / mod) * C (x % mod, y % mod) % mod;
}
ll f (ll x, ll y)
{
if (y < 0) return 0;
if (x <= y) return qwq (2, x);
if (x == 0) return 1;
if (y == 1) return x + 1;
int xp = x % mod, Xp = x / mod, yp = y % mod, Yp = y / mod, res = 0, ress = 0;
for (int i = 0; i < mod; ++ i) ress = (ress + C (xp, i)) % mod;
for (int i = 0; i <= yp; ++ i) res = (res + C (xp, i)) % mod;
return (ress * f (Xp, Yp - 1) % mod + res * luc (Xp, Yp) % mod) % mod;
}
mizuki main ()
{
freopen ("dimension.in", "r", stdin);
freopen ("dimension.out", "w", stdout);
in (n), in (m), in (mod);
n ++, m ++;
if (n <= m) ot ((qwq (2, n) - 1 + mod) % mod);
else
{
p[0] = 1;
for (int i = 1; i < mod; ++ i) p[i] = p[i - 1] * i % mod;
ny[mod - 1] = qwq (p[mod - 1], mod - 2);
for (int i = mod - 2; i >= 0; -- i) ny[i] = ny[i + 1] * (i + 1) % mod;
ot ((f (n, m) - 1 + mod) % mod);
}
return love;
}
无双挑战 (challenge)
非常非常J的题，考虑每个点的贡献，他的贡献是\(\frac{1}{n(n-1)}\sum\limits_{i = 1}^{n}\sum\limits_{J = 1}^{n}(m-a_i)*dis_{i, J}\) \((m-a_i)\)可以直接提到前面，那么只用考虑\(\sum\limits_{J = 1}^{n}dis_{i,J}\)
\(dis_{i, J} = dep_i +dep_J - 2\times dep_{lca_{i, J}}\)
对于\(dep_i\)和\(dep_J\)可以直接按\(a_i\)从大到小枚举，直接记录下来有多少比当前节点大的，只有这些点会产生贡献
那么只剩\(\sum\limits_{J=1}^{n}dep_{lca_{i,J}}\)了
设每个节点的子树内比当前节点大的点（包括自己）的数量为\(size_i\)
对于自己子树内的点，\(lca\)一定是自己，对于子树外的点\(lca\)一定是自己的祖先
所以\(\sum\limits_{J=0}^{n}dep_{lca_{i,J}}=size_i\times dep_i +(size_{fa_i} - size_i)\times dep_{fa_i}+\cdots +(size_1 - size_k)\times 1\)
整理可得\(\sum\limits_{J=0}^{n}dep_{lca_{i,J}}=size_i\times(dep_i-dep_{fa_i})+size_{fa_i}\times(dep_{fa_i} - dep_{fa_{fa_i}})+\cdots+size_1\times dep_1=size_i+size_{fa_i}+\cdots+size_1\)
于是就可以树剖搞一搞了，直接查从它到根上的\(size\)之和，当一个点计算完贡献后就给从他到根的路径上的每个点的\(size+1\)
注意因为要计算严格小于自己的，所以计算完一个点的贡献后要先存起来，等到下一个数比自己大的时候再一起进行修改
code
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0; char f = 1; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 0;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (!f) x = -x;
}
int stk[40], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
const int maxn = 8e5 + 10;
const int mod = 998244353;
#define int ll
ll n, m;
ll qwq (ll x, ll y)
{
ll res = 1;
while (y)
{
if (y & 1) res = res * x % mod;
x = x * x % mod;
y >>= 1;
}
return res;
}
struct miuna
{
int v, nt;
}e[maxn << 1];
int tot, head[maxn];
inline void add (int u, int v)
{
tot ++;
e[tot].v = v;
e[tot].nt = head[u];
head[u] = tot;
}
int dep[maxn], siz[maxn], fat[maxn], top[maxn], son[maxn], num[maxn], tim;
void dfs (int x, int fa)
{
dep[x] = dep[fa] + 1;
siz[x] = 1;
fat[x] = fa;
for (int i = head[x]; i; i = e[i].nt)
{
int y = e[i].v;
if (y == fa) continue;
dfs (y, x);
siz[x] += siz[y];
if (siz[y] > siz[son[x]]) son[x] = y;
}
}
void dfs2 (int x, int tp)
{
top[x] = tp;
num[x] = ++ tim;
if (son[x]) dfs2 (son[x], tp);
for (int i = head[x]; i; i = e[i].nt)
{
int y = e[i].v;
if (y == fat[x] || y == son[x]) continue;
dfs2 (y, y);
}
}
int t[maxn << 2], lz[maxn << 2];
inline void pushdown (int k, int l, int r)
{
int mid = (l + r) >> 1;
t [k << 1] += lz[k] * (mid - l + 1);
lz[k << 1] += lz[k];
t [k << 1 | 1] += lz[k] * (r - mid);
lz[k << 1 | 1] += lz[k];
lz[k] = 0;
}
inline void pushup (int k)
{
t[k] = t[k << 1] + t[k << 1 | 1];
}
int query (int k, int l, int r, int L, int R)
{
if (L <= l && r <= R) return t[k];
if (lz[k]) pushdown (k, l, r);
int mid = (l + r) >> 1;
int res = 0;
if (L <= mid) res += query (k << 1, l, mid, L, R);
if (R > mid) res += query (k << 1 | 1, mid + 1, r, L, R);
return res;
}
void update (int k, int l, int r, int L, int R)
{
if (L <= l && r <= R)
{
t[k] += r - l + 1;
lz[k] += 1;
return;
}
if (lz[k]) pushdown (k, l, r);
int mid = (l + r) >> 1;
if (L <= mid) update (k << 1, l, mid, L, R);
if (R > mid) update (k << 1 | 1, mid + 1, r, L, R);
pushup (k);
}
int q_ (int x, int y)
{
int ans = 0;
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]]) swap (x, y);
ans += query (1, 1, n, num[top[x]], num[x]);
x = fat[top[x]];
}
if (dep[x] > dep[y]) swap (x, y);
ans += query (1, 1, n, num[x], num[y]);
return ans;
}
void upd (int x, int y)
{
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]]) swap (x, y);
update (1, 1, n, num[top[x]], num[x]);
x = fat[top[x]];
}
if (dep[x] > dep[y]) swap (x, y);
update (1, 1, n, num[x], num[y]);
}
struct Shiodome
{
int a, id;
friend bool operator < (Shiodome a, Shiodome b)
{
return a.a > b.a;
}
}a[maxn];
int st[maxn], topp;
mizuki main ()
{
freopen ("challenge.in", "r", stdin);
freopen ("challenge.out", "w", stdout);
in (n), in (m);
for (int i = 1; i <= n; ++ i) in (a[i].a), a[i].id = i;
for (int i = 1, x, y; i < n; ++ i)
{
in (x), in (y);
add (x, y);
add (y, x);
}
dfs (1, 0);
dfs2 (1, 1);
ll ans = 0;
sort (a + 1, a + 1 + n);
ll sum1 = 0, sum2 = 0;
for (int i = 1; i <= n; ++ i)
{
if (i != 1 && a[i].a != a[i - 1].a) while (topp) upd (1, a[st[topp]].id), sum1 ++, sum2 += dep[a[st[topp --]].id];
st[++ topp] = i;
ans = (ans + (m - a[i].a) * ((sum1 * dep[a[i].id] % mod + sum2 - 2 * q_ (1, a[i].id) % mod) % mod + mod) % mod) % mod;
}
ot (ans * qwq (n, mod - 2) % mod * qwq (n - 1, mod - 2) % mod);
return love;
}
