A - Anti-Adjacency

K <= (N + 1) / 2

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 40005

#define eps 1e-12

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	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,K;

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	read(N);read(K);

	if(K <= (N + 1) / 2) puts("YES");

	else puts("NO");

}

B - Path

点度都不超过3

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 40005

#define eps 1e-12

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	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 deg[5];

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	int a,b;

	for(int i = 1 ; i <= 3 ; ++i) {

		read(a);read(b);

		++deg[a];++deg[b];

	}

	for(int i = 1 ; i <= 4 ; ++i) {

		if(deg[i] >= 3) {puts("NO");return 0;}

	}

	puts("YES");

}

C - When I hit my pocket...

先给自己加到A块饼干

然后判断一下B - A是否大于2，然后就用两次不断的换B - A

否则就直接一直拍得一次饼干就行

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 40005

#define eps 1e-12

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	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);

}

int64 K,A,B;

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	read(K);read(A);read(B);

	int64 now = 1;

	if(now < A) {

		int64 t = min(A - now,K);

		now += t;

		K -= t;

	}

	if(B - A > 2) {

		now += K / 2 * (B - A);

		K %= 2;

	}

	now += K;

	out(now);enter;

}

D - Ears

DEF我做题顺序是反的，D题我做的有点智障

FE做完之后一开榜19，D做完掉到45

大家AK水平那么高= =

事实上就是一个有点智障的dp

空 偶 奇 偶 空

每个段都可以为0，如果一个数在空段代价是它的值，奇数在偶段代价为1，偶数在奇数段代价为1，0在偶数段代价为2

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 200005

#define eps 1e-12

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	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 L;

int64 dp[MAXN][6],A[MAXN];

void Solve() {

	read(L);

	for(int i = 1 ; i <= L ; ++i) {read(A[i]);}

	for(int i = 1 ; i <= 4 ; ++i) dp[0][i] = 1e16;

	dp[0][0] = 0;

	for(int i = 1 ; i <= L ; ++i) {

		dp[i][0] = dp[i - 1][0] + A[i];

		int t = A[i] & 1;

		dp[i][1] = min(dp[i - 1][1],dp[i - 1][0]);

		dp[i][2] = min(min(dp[i - 1][1],dp[i - 1][0]),dp[i - 1][2]);

		if(A[i] % 2 == 0) dp[i][2]++;

		dp[i][3] = min(dp[i - 1][2],dp[i - 1][3]);

		if(A[i] & 1) {dp[i][1]++;dp[i][3]++;}

		if(A[i] == 0) {dp[i][1] += 2;dp[i][3] += 2;}

		dp[i][4] = min(min(dp[i - 1][3],dp[i - 1][2]),min(dp[i - 1][1],dp[i - 1][4]));

		dp[i][4] += A[i];

	}

	int64 res = dp[L][0];

	for(int i = 1 ; i <= 4 ; ++i) res = min(res,dp[L][i]);

	out(res);enter;

}

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	Solve();

}

E - Odd Subrectangles

如果选的行数确定了，那么选某一列的奇偶性也会确定，如果奇数有a列，偶数有b列，\(a + b = M\)

那么方案数是选a中奇数个，b随便选

方案数是\(2^{a - 1}\cdot 2^{b}\)，发现这个就是\(2^{M - 1}\)

所以只要存在a即可，即选的行数的数异或起来值不为0，转而求异或为0的方案，直接上线性基即可

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 40005

#define eps 1e-12

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	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 = 998244353;

int N,M;

int a[305][305];

int b[305][305],cnt;

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 update(int &x,int y) {

	x = inc(x,y);

}

int fpow(int x,int c) {

	int res = 1,t = x;

	while(c) {

		if(c & 1) res = mul(res,t);

		t = mul(t,t);

		c >>= 1;

	}

	return res;

}

void Solve() {

	read(N);read(M);

	for(int i = 1 ; i <= N ; ++i) {

		for(int j = 1 ; j <= M ; ++j) {

			read(a[i][j]);

		}

	}

	for(int i = 1 ; i <= N ; ++i) {

		for(int j = 1 ; j <= M ; ++j) {

			if(a[i][j]) {

				if(!b[j][j]) {

					for(int k = 1 ; k <= M ; ++k) b[j][k] = a[i][k];

					++cnt;

					break;

				}

				else {

					for(int k = 1 ; k <= M ; ++k) a[i][k] ^= b[j][k];

				}

			}

		}

	}

	int ans = inc(fpow(2,N),MOD - fpow(2,N - cnt));

	ans = mul(ans,fpow(2,M - 1));

	out(ans);enter;

}

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	Solve();

}

F - Pass

直接转二维平面，然后第i步能走到的点\((a,b)\)a个红球，b个蓝球，要满足a小于等于前i个人红球总和，b小于等于前i个人蓝球总和

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 40005

#define eps 1e-12

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	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 = 998244353;

char s[2005];

int sum[2005][2],N;

int dp[4005][4005];

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 update(int &x,int y) {

	x = inc(x,y);

}

void Solve() {

	scanf("%s",s + 1);

	N = strlen(s + 1);

	for(int i = 1 ; i <= N ; ++i) {

		sum[i][0] = sum[i - 1][0];

		sum[i][1] = sum[i - 1][1];

		if(s[i] == '0') sum[i][0] += 2;

		if(s[i] == '1') {sum[i][0]++;sum[i][1]++;}

		if(s[i] == '2') sum[i][1] += 2;

	}

	dp[0][0] = 1;

	for(int i = 1 ; i <= 2 * N ; ++i) {

		int t = min(i,N);

		for(int j = 0 ; j <= i ; ++j) {

			int a = j,b = i - j;

			if(a <= sum[t][0] && b <= sum[t][1]) {

				if(a) update(dp[a][b],dp[a - 1][b]);

				if(b) update(dp[a][b],dp[a][b - 1]);

			}

		}

	}

	out(dp[sum[N][0]][sum[N][1]]);enter;

}

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	Solve();

}