第一题略
第二题组合数学
s1 = min(cnt['a'],cnt['b']),
s2 = min(cnt['c'],cnt['d']),
b1 = max(cnt['a'],cnt['b']),
b2 = max(cnt['c'],cnt['d']);
简单分开枚举'a','b'一组和'c','d'一组然后乘起来就好了
ans = sigma(C(s1,i) * C(b1,i))(0<=i<=s1) * sigma(C(s2,i) * C(b2,i))(0<=i<=s2) - 1;
#include<bits/stdc++.h>
#define MP make_pair
using namespace std;
typedef long long LL;
const int N = 5e5 + ;
const LL mod = 1e9 + ;
int Tm[];
char s[N];
LL fac[N];
LL inv[N];
LL qpow(LL x,LL b){
LL res = ;
while(b){
if(b&)res = res * x % mod;
x = x * x % mod;
b >>= ;
}
return res;
}
void init(){
fac[] = ;
for(int i = ; i < N; i++)fac[i] = fac[i-] * i % mod;
inv[N - ] = qpow(fac[N - ],mod - );
for(int i = N - ; i >= ; i--){
inv[i] = inv[i + ] * (i + ) % mod;
}
}
inline LL C(int a,int b){
return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
int s1,b1,s2,b2;
LL solve(){
LL res1 = ,res2 = ;
for(int i = ; i <= s1; i++){
res1 = (res1 + C(s1,i) * C(b1,i) % mod) % mod;
}
for(int i = ; i <= s2; i++){
res2 = (res2 + C(s2,i) * C(b2,i) % mod) % mod;
}
// cout<<s1<<" "<<s2<<endl;
// cout<<res1<<" "<<res2<<endl;
return (res1 * res2 % mod - + mod)%mod;
}
int main() {
// freopen("input", "r", stdin);
// freopen("input", "w", stdout);
ios::sync_with_stdio(false);
cin.tie();
init();
int q;
cin>>q;
while(q--){
memset(Tm,,sizeof Tm);
cin>>s;
int len = strlen(s);
for(int i = ; i < len; i++)Tm[(int)s[i]]++;
s1 = min(Tm['a'],Tm['b']);
b1 = max(Tm['a'],Tm['b']);
s2 = min(Tm['c'],Tm['d']);
b2 = max(Tm['c'],Tm['d']);
cout<<solve()<<endl;
}
return ; }
第三题对于每个询问建一颗虚树,根节点要放在虚树里,如果u , fa,fa是u的祖先,并且u到fa这条路径上只有这两个点,那这条路径的点子树黑点都等于u上的子树黑点。
#include<bits/stdc++.h>
#define MP make_pair
#define PB push_back
using namespace std;
typedef long long LL;
const int N = 6e5 + ;
const LL mod = 1e9 + ;
int L[N];
int n;
int R[N];
int ans[N];
int Fa[N][];
int deep[N];
int tin = ;
vector<int>Tree[N];
vector<int>Mtree[N];
int rev[N];
bool cmp(int a, int b) {
return L[a] < L[b];
}
int lca(int a, int b) {
if(deep[a] < deep[b])swap(a, b);
int d = deep[a] - deep[b];
for(int i = ; i < ; i++) {
if(d & ( << i))a = Fa[a][i];
}
if(a == b)return a;
for(int i = ; i >= ; i--) {
if(Fa[a][i] != Fa[b][i])a = Fa[a][i], b = Fa[b][i];
}
return Fa[a][];
}
int v[N];
int St[N];
int top = ;
inline bool isFa(int fa,int u){
return L[fa] <= L[u] && R[fa] >= R[u];
}
inline void dfs(int u,int fa){
for(int to : Mtree[u]){
dfs(to,u);
v[u] += v[to];
}
ans[v[u]] += deep[u] - deep[fa];
}
inline void solve() {
int k, x;
cin >> k;
int tol = ;
for(int i = ; i < k; i++) {
cin >> x;
rev[tol++] = x;
v[x] = ;
}
sort(rev, rev + tol, cmp);
for(int i = ; i < k; i++) {
rev[tol++] = lca(rev[i - ], rev[i]);
}
rev[tol++] = ;
sort(rev, rev + tol, cmp);
tol = unique(rev , rev + tol) - rev;
top = ;
St[++top] = ;
for(int i = ; i < tol; i++){
int to = rev[i];
while(top > && !isFa(St[top],to))top--;
int nt = St[top];
Mtree[nt].PB(to);
St[++top] = to;
}
dfs(,);
ans[] = n;
for(int i = ; i <= k; i++)ans[] -= ans[i];
for(int i = ; i <= k; i++)cout<<ans[i]<<" ",ans[i] = ;
for(int i = ; i < tol; i++){
Mtree[rev[i]].clear();v[rev[i]] = ;
}
cout<<endl;
}
void pre(int u, int fa) {
deep[u] = deep[fa] + ;
L[u] = ++tin;
Fa[u][] = fa;
for(int to : Tree[u]) {
if(to == fa)continue;
pre(to, u);
}
R[u] = tin;
}
void init() {
cin>>n;
for(int i = ; i < n; i++) {
int u, to;
cin >> u >> to;
Tree[u].PB(to), Tree[to].PB(u);
}
pre(, );
for(int i = ; ( << i) < n; i++) {
for(int j = ; j <= n; j++) {
Fa[j][i] = Fa[Fa[j][i - ]][i - ];
}
}
int q;
cin >> q;
while(q--) {
solve();
}
}
int main() {
// freopen("input", "r", stdin);
// freopen("input", "w", stdout);
ios::sync_with_stdio(false);
cin.tie();
init();
return ;
}