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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#408417 | #4811. Be Careful | chenxinyang2006 | WA | 4ms | 38912kb | C++20 | 8.2kb | 2024-05-10 11:10:21 | 2024-05-10 11:10:22 |
Judging History
answer
#include <bits/stdc++.h>
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
#define mod 998244353
//#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;
template <class T>
void chkmax(T &x,T y){
if(x < y) x = y;
}
template <class T>
void chkmin(T &x,T y){
if(x > y) x = y;
}
inline int popcnt(int x){
return __builtin_popcount(x);
}
inline int ctz(int x){
return __builtin_ctz(x);
}
template <int P>
class mod_int
{
using Z = mod_int;
private:
static int mo(int x) { return x < 0 ? x + P : x; }
public:
int x;
int val() const { return x; }
mod_int() : x(0) {}
template <class T>
mod_int(const T &x_) : x(x_ >= 0 && x_ < P ? static_cast<int>(x_) : mo(static_cast<int>(x_ % P))) {}
bool operator==(const Z &rhs) const { return x == rhs.x; }
bool operator!=(const Z &rhs) const { return x != rhs.x; }
Z operator-() const { return Z(x ? P - x : 0); }
Z pow(long long k) const
{
Z res = 1, t = *this;
while (k)
{
if (k & 1)
res *= t;
if (k >>= 1)
t *= t;
}
return res;
}
Z &operator++()
{
x < P - 1 ? ++x : x = 0;
return *this;
}
Z &operator--()
{
x ? --x : x = P - 1;
return *this;
}
Z operator++(int)
{
Z ret = x;
x < P - 1 ? ++x : x = 0;
return ret;
}
Z operator--(int)
{
Z ret = x;
x ? --x : x = P - 1;
return ret;
}
Z inv() const { return pow(P - 2); }
Z &operator+=(const Z &rhs)
{
(x += rhs.x) >= P && (x -= P);
return *this;
}
Z &operator-=(const Z &rhs)
{
(x -= rhs.x) < 0 && (x += P);
return *this;
}
Z operator-()
{
return -x;
}
Z &operator*=(const Z &rhs)
{
x = 1ULL * x * rhs.x % P;
return *this;
}
Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
#define setO(T, o) \
friend T operator o(const Z &lhs, const Z &rhs) \
{ \
Z res = lhs; \
return res o## = rhs; \
}
setO(Z, +) setO(Z, -) setO(Z, *) setO(Z, /)
#undef setO
friend istream& operator>>(istream& is, mod_int& x)
{
long long tmp;
is >> tmp;
x = tmp;
return is;
}
friend ostream& operator<<(ostream& os, const mod_int& x)
{
os << x.val();
return os;
}
};
using Z = mod_int<mod>;
Z power(Z p,ll k){
Z ans = 1;
while(k){
if(k % 2 == 1) ans *= p;
p *= p;
k /= 2;
}
return ans;
}
int n;
vector <int> G[205];
int sz[205];
Z cof[205][205],C[205][205],P[205][205];//cof[i][j],i 个 [0,n] 整数,要求有 j 种值出现 >= 1 次
Z val[205][205],sval[205][205],f[1 << 21],g[1 << 21];
Z comb[1 << 21],prd[1 << 21];
vector <Z> dp[1 << 21],ndp[1 << 21];
int _l,idx[205];
void dfs(int u,int fa){
for(int v:G[u]){
if(v == fa) continue;
dfs(v,u);
sz[u] += sz[v] + (SZ(G[v]) == 1);
}
if(SZ(G[u]) == 1) return;
int B = inf,l = inf,qwq,leaf = 0;
rep(b,0,22){
qwq = 0;
for(int v:G[u]) if(v != fa && SZ(G[v]) > max(1,b)) qwq++;
if(b + qwq < B + l){
B = b;
l = qwq;
}
}
fill(f,f + (1 << B),0);
f[0] = 1;
for(int v:G[u]){
if(v == fa) continue;
if(SZ(G[v]) == 1){
leaf++;
continue;
}else if(SZ(G[v]) > B){
continue;
}
rep(i,0,SZ(G[v]) - 1){
rep(S,0,(1 << B) - 1) g[S | (1 << i)] += f[S] * val[v][i];
}
rep(S,0,(1 << B) - 1){
f[S] = g[S];
g[S] = 0;
}
}
// printf("slv %d B=%d l=%d\n",u,B,l);
// rep(S,0,(1 << B) - 1) printf("%d ",f[S].val());
// printf("\n");
_l = 0;
for(int v:G[u]) if(v != fa && SZ(G[v]) > max(1,B)) idx[_l++] = v;
// rep(i,0,_l - 1) printf("%d ",idx[i]);
// printf("\n");
rep(S,0,(1 << (B + l)) - 1){
dp[S].resize(1);
dp[S][0] = 0;
}
rep(S,0,(1 << B) - 1) dp[S << l][0] = f[S];
//[0,B-1] old,[B,B+l-1] new
rep(i,0,B - 1){
rep(s,0,l - 1){
rep(_S,0,(1 << (B - i)) - 1){
rep(SS,_S << l,((_S + 1) << l) - 1){
if((SS >> s) & 1) continue;
rep(k,0,i) dp[SS | (1 << l) | (1 << s)][k] += dp[SS][k] * val[idx[s]][i];
}
}
}
rep(S,0,(1 << (l + B - i - 1)) - 1){
ndp[S].resize(i + 2);
dp[S].resize(i + 2);
}
rep(_S,0,(1 << (B - i)) - 1){
rep(SS,0,(1 << l) - 1){
if(_S & 1) rep(k,0,i) ndp[(_S >> 1) << l | SS][k] += dp[(_S << l) | SS][k];
else rep(k,0,i) ndp[(_S >> 1) << l | SS][k + 1] += dp[(_S << l) | SS][k];
}
}
rep(S,0,(1 << (l + B - i - 1)) - 1){
copy(ndp[S].begin(),ndp[S].begin() + i + 2,dp[S].begin());
fill(ndp[S].begin(),ndp[S].begin() + i + 2,0);
rep(k,0,i + 1) comb[S & ((1 << l) - 1)] += dp[S][k] * cof[leaf][k];
}
prd[0] = 1;
rep(S,1,(1 << l) - 1){
int p = ctz(S);
prd[S] = prd[S - (1 << p)] * sval[idx[p]][i + 1];
}
rep(S,0,(1 << l) - 1){
val[u][i] += comb[S] * prd[(1 << l) - 1 - S];
comb[S] = 0;
}
}
rep(S,0,(1 << l) - 1){
dp[S].resize(leaf + 1);
ndp[S].resize(leaf + 1);
// rep(k,0,leaf) printf("%d ",dp[S][k].val());
// printf("\n");
}
// printf("\n");
rep(i,B,SZ(G[u]) - 1){
rep(s,0,l - 1){
rep(S,0,(1 << l) - 1){
if((S >> s) & 1) continue;
rep(k,0,leaf) ndp[S + (1 << s)][k] += (dp[S][k] + ndp[S][k]) * val[idx[s]][i];
}
}
prd[0] = 1;
rep(S,1,(1 << l) - 1){
int p = ctz(S);
prd[S] = prd[S - (1 << p)] * sval[idx[p]][i + 1];
}
// printf("state %d\n",i);
rep(S,0,(1 << l) - 1){
per(k,leaf,1) dp[S][k] = dp[S][k - 1];
dp[S][0] = 0;
rep(k,0,leaf){
dp[S][k] += ndp[S][k];
val[u][i] += dp[S][k] * prd[(1 << l) - 1 - S] * cof[leaf][k];
ndp[S][k] = 0;
}
// rep(k,0,leaf) printf("%d ",dp[S][k].val());
// printf("\n");
}
}
/* printf("limit info of %d\n",u);
rep(i,0,SZ(G[u]) - 1) printf("%d ",val[u][i].val());
printf("\n");*/
per(i,SZ(G[u]),1) val[u][i] = val[u][i - 1] - val[u][i];
val[u][0] = P[n + 1][sz[u]] - val[u][0];
per(i,SZ(G[u]),0) sval[u][i] = sval[u][i + 1] + val[u][i];
/* printf("dp info of %d\n",u);
rep(i,0,SZ(G[u]) - 1) printf("%d ",val[u][i].val());
printf("\n");*/
//sval suf sum of val
}
int main(){
// freopen("test.in","r",stdin);
scanf("%d",&n);
rep(i,0,n + 1){
P[i][0] = 1;
rep(j,1,n) P[i][j] = P[i][j - 1] * i;
}
rep(i,0,n){
C[i][0] = 1;
rep(j,1,i) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
rep(i,0,n){
rep(j,0,n){
rep(k,0,j){
if(k & 1) cof[i][j] -= C[j][k] * P[n + 1 - k][i];
else cof[i][j] += C[j][k] * P[n + 1 - k][i];
}
}
}
rep(i,1,n - 1){
int u,v;
scanf("%d%d",&u,&v);
G[u].eb(v);G[v].eb(u);
}
dfs(1,0);
rep(i,0,n) printf("%d\n",val[1][i].val());
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 4ms
memory: 37348kb
input:
5 1 2 1 3 2 4 2 5
output:
55 127 34 0 0 0
result:
ok 6 numbers
Test #2:
score: 0
Accepted
time: 4ms
memory: 38548kb
input:
8 1 2 1 3 1 4 1 5 1 6 6 7 6 8
output:
69632 265534 133905 47790 12636 1944 0 0 0
result:
ok 9 numbers
Test #3:
score: -100
Wrong Answer
time: 4ms
memory: 38912kb
input:
3 1 2 2 3
output:
0 0 0 0
result:
wrong answer 1st numbers differ - expected: '1', found: '0'