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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#187935 | #7509. 01tree | bulijiojiodibuliduo# | TL | 13ms | 8916kb | C++17 | 4.3kb | 2023-09-25 06:57:12 | 2023-09-25 06:57:12 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef basic_string<int> BI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(1);
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
template<int MOD, int RT> struct mint {
static const int mod = MOD;
static constexpr mint rt() { return RT; } // primitive root for FFT
int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
mint():v(0) {}
mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
if (v < 0) v += MOD; }
bool operator==(const mint& o) const {
return v == o.v; }
friend bool operator!=(const mint& a, const mint& b) {
return !(a == b); }
friend bool operator<(const mint& a, const mint& b) {
return a.v < b.v; }
mint& operator+=(const mint& o) {
if ((v += o.v) >= MOD) v -= MOD;
return *this; }
mint& operator-=(const mint& o) {
if ((v -= o.v) < 0) v += MOD;
return *this; }
mint& operator*=(const mint& o) {
v = int((ll)v*o.v%MOD); return *this; }
mint& operator/=(const mint& o) { return (*this) *= inv(o); }
friend mint pow(mint a, ll p) {
mint ans = 1; assert(p >= 0);
for (; p; p /= 2, a *= a) if (p&1) ans *= a;
return ans; }
friend mint inv(const mint& a) { assert(a.v != 0);
return pow(a,MOD-2); }
mint operator-() const { return mint(-v); }
mint& operator++() { return *this += 1; }
mint& operator--() { return *this -= 1; }
friend mint operator+(mint a, const mint& b) { return a += b; }
friend mint operator-(mint a, const mint& b) { return a -= b; }
friend mint operator*(mint a, const mint& b) { return a *= b; }
friend mint operator/(mint a, const mint& b) { return a /= b; }
};
const int MOD=1000000007;
using mi = mint<MOD,5>; // 5 is primitive root for both common mods
namespace simp {
vector<mi> fac,ifac,invn;
void check(int x) {
if (fac.empty()) {
fac={mi(1),mi(1)};
ifac={mi(1),mi(1)};
invn={mi(0),mi(1)};
}
while (SZ(fac)<=x) {
int n=SZ(fac),m=SZ(fac)*2;
fac.resize(m);
ifac.resize(m);
invn.resize(m);
for (int i=n;i<m;i++) {
fac[i]=fac[i-1]*mi(i);
invn[i]=mi(MOD-MOD/i)*invn[MOD%i];
ifac[i]=ifac[i-1]*invn[i];
}
}
}
mi gfac(int x) {
check(x); return fac[x];
}
mi ginv(int x) {
check(x); return invn[x];
}
mi gifac(int x) {
check(x); return ifac[x];
}
mi binom(int n,int m) {
if (m < 0 || m > n) return mi(0);
return gfac(n)*gifac(m)*gifac(n - m);
}
}
const int N=101000;
VI e[N];
int par[N],c0[N],cm[N],d0[N],dm[N],a[N],b[N];
int a0,am,b0,bm,n;
char s1[N],s2[N];
mi ans;
void dfs0(int u,int f) {
for (auto v:e[u]) if (v!=f) {
par[v]=par[u]^1;
dfs0(v,u);
}
}
void dfs(int u,int f) {
c0[u]=a[u]==0,cm[u]=a[u]==-1;
d0[u]=b[u]==0,dm[u]=b[u]==-1;
for (auto v:e[u]) if (v!=f) {
dfs(v,u);
c0[u]+=c0[v];
cm[u]+=cm[v];
d0[u]+=d0[v];
dm[u]+=dm[v];
}
int c1=c0[u],c2=cm[u],c3=a0-c1,c4=am-c2;
int d1=d0[u],d2=dm[u],d3=b0-d1,d4=bm-d2;
rep(i1,0,c2+1) rep(i2,0,c4+1) rep(j1,0,d2+1) rep(j2,0,d4+1) {
if (c1+c3+i1+i2==d1+d3+j1+j2) {
ans+=simp::binom(c2,i1)*simp::binom(d2,j1)*simp::binom(c4,i2)
*simp::binom(d4,j2)*mi(abs((c1+i1)-(d1+j1)));
}
}
}
void solve() {
scanf("%d",&n);
rep(i,1,n+1) e[i].clear();
rep(i,1,n) {
int u,v;
scanf("%d%d",&u,&v);
e[u].pb(v);
e[v].pb(u);
}
dfs0(1,0);
scanf("%s",s1+1);
rep(i,1,n+1) {
a[i]=s1[i]=='?'?-1:(s1[i]-'0');
if (a[i]!=-1) a[i]^=par[i];
}
scanf("%s",s2+1);
rep(i,1,n+1) {
b[i]=s2[i]=='?'?-1:(s2[i]-'0');
if (b[i]!=-1) b[i]^=par[i];
}
a0=0,am=0;
b0=0,bm=0;
rep(i,1,n+1) {
a0+=(a[i]==0),am+=(a[i]==-1);
b0+=(b[i]==0),bm+=(b[i]==-1);
}
ans=0;
dfs(1,0);
printf("%d\n",(int)ans);
}
int _;
int main() {
for (scanf("%d",&_);_;_--) {
solve();
}
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 8916kb
input:
3 2 1 2 00 11 3 1 2 2 3 ??? ??? 3 1 2 2 3 ??1 0?0
output:
1 16 1
result:
ok 3 number(s): "1 16 1"
Test #2:
score: 0
Accepted
time: 13ms
memory: 7292kb
input:
1000 23 1 2 1 3 1 4 2 5 5 6 4 7 3 8 4 9 8 10 8 11 8 12 1 13 7 14 10 15 7 16 7 17 5 18 18 19 12 20 9 21 21 22 6 23 00?10?0000??1?00111?010 011??1?10?01?110?0??101 6 1 2 1 3 1 4 4 5 3 6 000?10 1???01 25 1 2 2 3 2 4 4 5 5 6 2 7 4 8 5 9 7 10 8 11 11 12 5 13 11 14 3 15 6 16 14 17 1 18 4 19 6 20 4 21 5 22...
output:
53545 24 11724 2228 162 14 711 28 550 1680 52 2 13 988 9480 2342 626416 0 71780 1 88 39207 19448 4 37395 9602 3253496 38057200 1066 3 303732 1608 281022 11718 170 78 15 1219376 29364 9092 7 0 2 7014 1942448 170371 75845 167217 16554 64 904 564290 14822 1127090 1758504 1227646 14833316 14954550 36055...
result:
ok 1000 numbers
Test #3:
score: -100
Time Limit Exceeded
input:
1 3000 1 2 2 3 2 4 1 5 3 6 2 7 5 8 8 9 9 10 10 11 2 12 11 13 7 14 11 15 7 16 13 17 8 18 1 19 11 20 10 21 18 22 7 23 7 24 15 25 23 26 24 27 14 28 15 29 25 30 16 31 6 32 10 33 3 34 30 35 16 36 9 37 36 38 28 39 26 40 33 41 1 42 11 43 20 44 23 45 14 46 31 47 41 48 11 49 48 50 45 51 6 52 10 53 32 54 38 5...