C - Good Sequence
题解
用个map愉悦一下就好了
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
int a[MAXN],ans;
map<int,int> zz;
void Init() {
read(N);
for(int i = 1 ; i <= N ; ++i) {read(a[i]);zz[a[i]] += 1;}
sort(a + 1,a + N + 1);
N = unique(a + 1,a + N + 1) - a - 1;
}
void Solve() {
for(int i = 1 ; i <= N ; ++i) {
if(zz[a[i]] < a[i]) ans += zz[a[i]];
else ans += zz[a[i]] - a[i];
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}
D - FT Robot
如果经过了奇数个T，那么原来在x轴上走，后来肯定在y轴上走，原来y轴变为x轴
如果偶数个T，那么保持原来的轴不变
我们会选定一个方向使得在这个方向上走后面的F的个数
这个可以用一个简单的背包（加上bitset优化一下）实现，注意在序列最前的F手动走完，因为无法改成X轴负方向
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,x,y,s,t;
char str[8005];
vector<int> dir[2];
bitset<16005> f[2];
bool cmp(int a,int b) {
return a > b;
}
void Solve() {
scanf("%s",str + 1);read(x);read(y);
N = strlen(str + 1);
int d = 0;
s = 0,t = 0;
int p = 1;
while(p <= N && str[p] == 'F') {++s;++p;}
int cnt = 0;
for(int i = p ; i <= N ; ++i) {
if(str[i] == 'T') {
if(cnt) dir[d].pb(cnt);
d ^= 1;cnt = 0;
}
else {++cnt;}
}
if(cnt) dir[d].pb(cnt);
if(abs(s - x) > N) {puts("No");return;}
int cur = 0;
f[cur].reset();
f[cur][8000] = 1;
for(auto k : dir[0]) {
f[cur ^ 1].reset();
f[cur ^ 1] = (f[cur] << k) | (f[cur] >> k);
cur ^= 1;
}
if(!f[cur][8000 + abs(s - x)]) {puts("No");return;}
cur = 0;
f[cur].reset();
f[cur][8000] = 1;
for(auto k : dir[1]) {
f[cur ^ 1].reset();
f[cur ^ 1] = (f[cur] << k) | (f[cur] >> k);
cur ^= 1;
}
if(!f[cur][8000 + abs(t - y)]) {puts("No");return;}
puts("Yes");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E-Prefix-free Game
给每个可以走的点的sg函数简单推一下，从第(L - 1)层开始归纳
很容易发现sg函数是lowbit(L - dep)
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int son[MAXN][2],N,rt = 1,Ncnt = 1;
int64 L,dep[MAXN];
char s[MAXN];
int64 lowbit(int64 x) {
return x & -x;
}
void Insert(char *str) {
int len = strlen(str + 1);
int p = rt;
for(int i = 1 ; i <= len ; ++i) {
if(!son[p][str[i] - '0']) {
son[p][str[i] - '0'] = ++Ncnt;
dep[Ncnt] = dep[p] + 1;
}
p = son[p][str[i] - '0'];
}
}
void Solve() {
read(N);read(L);
for(int i = 1 ; i <= N ; ++i) {
scanf("%s",s + 1);
Insert(s);
}
int64 ans = 0;
for(int i = 1 ; i <= Ncnt ; ++i) {
if((son[i][0] == 0 && son[i][1] != 0) || (son[i][0] != 0 && son[i][1] == 0)) {
ans ^= lowbit(L - dep[i]);
}
}
if(ans) puts("Alice");
else puts("Bob");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Squirrel Migration
第一个结论就推不出来QAQ
就是考虑一条边，边断掉分成两个子树是(x,y)，这条边最多被经过(2min{siz[x],siz[y]})
考虑有两个重心（也就是断开一条边可以分成两个大小为(N / 2)的子树）
肯定可以构造合法的解，方案数就是((frac{N}{2}!)^2)
有一个重心也是有解的，就是把所有的儿子从大到小排序
把最大的放在第一个，然后把剩下从小到大放，重心到它自己即可
这是一个合法解
那么我们可以知道，满足条件的答案一定是重心随便放，断掉重心后，剩下的点必须不选自己所在的树里的点
我们可以容斥做，用(f[k])选了k个点，且这k个点能到达的点都是自己子树中的点，类似树背包那样转移
答案就是(sum_{i = 0}^{N - 1} (-1)^{i}f[i] cdot (N - i)!)
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N;
struct node {
int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,siz[MAXN],son[MAXN];
int C[MAXN][MAXN],fac[MAXN],val[MAXN],tot,f[MAXN];
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void add(int u,int v) {
E[++sumE].next = head[u];
E[sumE].to = v;
head[u] = sumE;
}
void dfs(int u,int fa) {
siz[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
dfs(v,u);
siz[u] += siz[v];
son[u] = max(son[u],siz[v]);
}
}
son[u] = max(son[u],N - siz[u]);
}
void Init() {
read(N);
int x,y;
for(int i = 1 ; i < N ; ++i) {
read(x);read(y);
add(x,y);add(y,x);
}
C[0][0] = 1;
for(int i = 1 ; i <= N ; ++i) {
C[i][0] = 1;
for(int j = 1 ; j <= N ; ++j) {
C[i][j] = inc(C[i - 1][j],C[i - 1][j - 1]);
}
}
fac[0] = 1;
for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
}
void Solve() {
dfs(1,0);
int g = 1;
for(int i = 2 ; i <= N ; ++i) {
if(son[i] < son[g]) g = i;
}
if(N % 2 == 0 && son[g] == N / 2) {
out(mul(fac[N / 2],fac[N / 2]));enter;return;
}
for(int i = head[g] ; i ; i = E[i].next) {
int v = E[i].to;
if(siz[v] < siz[g]) val[++tot] = siz[v];
}
if(g != 1) val[++tot] = N - siz[g];
int s = 0;f[0] = 1;
for(int i = 1 ; i <= tot ; ++i) {
for(int t = s + val[i] ; t >= 0 ; --t) {
for(int j = 1 ; j <= val[i] ; ++j) {
if(j > t) break;
f[t] = inc(f[t],mul(mul(f[t - j],C[val[i]][j]),mul(C[val[i]][j],fac[j])));
}
}
s += val[i];
}
int t = 1,ans = 0;
for(int i = 0 ; i <= N - 1; ++i) {
ans = inc(ans,mul(mul(t,f[i]),fac[N - i]));
t = mul(t,MOD - 1);
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}