【AtCoder】CODE FESTIVAL 2017 qual A

A - Snuke's favorite YAKINIKU

……

#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);
}
string s;
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	cin >> s;
	if(s.substr(0,4) == "YAKI") puts("Yes");
	else puts("No");
}

B - fLIP

枚举N有几个按了，M有几列按了

对于一个点，行和列只有一个按了，那么这个点是黑的

#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,M,K;
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	read(N);read(M);read(K);
	for(int i = 0 ; i <= N ; ++i) {
		for(int j = 0 ; j <= M ; ++j) {
			int res = i * (M - j) + (N - i) * j;
			if(res == K) {puts("Yes");return 0;}
		}
	}
	puts("No");
	return 0;
}

C - Palindromic Matrix

从最外一圈开始推，我们输出需要几个4个一样的，2个一样的，可能还有需要1个单个的

然后用每个字母开始填4个一样的，再填2个一样的，再填1个单个的

如果没填完，那么就构造不出来

#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,M;
int num[30],cnt[10];
char s[105][105];

int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	read(N);read(M);
	for(int i = 1 ; i <= N ; ++i) {
		scanf("%s",s[i] + 1);
	}
	for(int i = 1 ; i <= N ; ++i) {
		for(int j = 1 ; j <= M ; ++j) {
			num[s[i][j] - 'a']++;
		}
	}
	int lr = 1,rr = N,lc = 1,rc = M;
	while(lr <= rr && lc <= rc) {
		int r = 0;
		if(lr != rr) r += 2;
		if(lc != rc) r += 2;
		if(!r) r = 1;
		cnt[r]++;
	 	for(int i = 1 ; i <= N ; ++i) {
			if(lr + i <= rr - i) {
				int r = 0;
				if(lr + i != rr - i) r += 2;
				if(lc != rc) r += 2;
				cnt[r]++;
			}
			else break;
		}
		for(int i = 1 ; i <= M ; ++i) {
			if(lc + i <= rc - i) {
				int r = 0;
				if(lc + i != rc - i) r += 2;
				if(rr != lr) r += 2;
				cnt[r]++;
			}
			else break;
		}
		++lr;--rr;++lc;--rc;
	}
	for(int i = 0 ; i < 26 ; ++i) {
		while(cnt[4] && num[i] >= 4) {num[i] -= 4;--cnt[4];}
		while(cnt[2] && num[i] >= 2) {num[i] -= 2;--cnt[2];}
		while(cnt[1] && num[i] >= 1) {num[i] -= 1;--cnt[1];}
	}
	if(cnt[1] || cnt[2] || cnt[4]) puts("No");
	else puts("Yes");
	return 0;
}

D - Four Coloring

把图旋转45度后，再放一个每块大小是d*d的方格，然后一行用RB间隔染色，下一行用GY间隔染色

#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 H,W,d;
bool vis[1005][1005];
string s = "RBGY";
void Solve() {
	read(H);read(W);read(d);
	for(int i = 1 ; i <= H ; ++i) {
		for(int j = 1 ; j <= W ; ++j) {
			int t = 0;
			if((i + j - 1) / d & 1) t |= 2;
			if((i - j + 500 - 1) / d & 1) t |= 1;
			putchar(s[t]);
		}
		enter;
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

E - Modern Painting

第一次肯定是一个竖直的完整画下来，或者一个水平的完整的画下来

用竖直的举例

就是枚举一个区间\([L,R]\)都是完整的竖直画下来，方案数是2的这个区间上下都有人的列个数次幂

然后算前后的那一块方案数

肯定是先横着一刀，再竖着一刀，再横着一刀

我们把横着的轮廓线画出来，再把一边的翻过去，发现是一个路径计数，但是要求对称轴必须有一个竖线

加入竖直有X个人，上有Y人，下有Z人，那么如果直接走过去，方案数是

\(\binom{X + Y + Z}{X}\)

那我们考虑一个序列，我们就走\(X - 1\)个竖直的，横向走到对称轴后，紧接着就添加一个竖直的

方案数就是

\(\binom{X + Y + Z - 1}{X - 1}\)就是方案数

#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 200500
#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;
char a[4][MAXN];
int fac[MAXN * 2],invfac[MAXN * 2],f[MAXN],b[MAXN],sum[4][MAXN],s[MAXN];
int pw[MAXN],ipw[MAXN],ans;
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;
}
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 update(int &x,int y) {
	x = inc(x,y);
}
int C(int n,int m) {
	if(n < m) return 0;
	return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void CalcV() {
	memset(f,0,sizeof(f));
	memset(b,0,sizeof(b));
	memset(s,0,sizeof(s));
	int X = sum[0][N],Y,Z;
	for(int i = 0 ; i <= M ; ++i) {
		if(a[2][i + 1] == '0' && a[3][i + 1] == '0') continue;
		Y = sum[2][i],Z = sum[3][i];
		if(!X) {
			if(Y == 0 && Z == 0) f[i] = 1;
			else f[i] = 0;
		}
		else {
			f[i] = C(X + Y + Z - 1,X - 1);
		}
	}
	X = sum[1][N];
	for(int i = M + 1 ; i >= 1 ; --i) {
		if(a[2][i - 1] == '0' && a[3][i - 1] == '0') continue;
	 	Y = sum[2][M] - sum[2][i - 1],Z = sum[3][M] - sum[3][i - 1];
		if(!X) {
			if(Y == 0 && Z == 0) b[i] = 1;
			else b[i] = 0;
		}
		else {
			b[i] = C(X + Y + Z - 1,X - 1);
		}
	}
	for(int i = 1 ; i <= M ; ++i) s[i] = s[i - 1] + (a[2][i] == '1' && a[3][i] == '1');
	for(int i = 0 ; i <= M ; ++i) {
		f[i] = mul(f[i],ipw[s[i]]);
	}
	for(int i = M + 1 ; i >= 1 ; --i) {
		b[i] = inc(b[i + 1],mul(b[i],pw[s[i - 1]]));
	}
	for(int i = 1 ; i <= M ; ++i) {
		update(ans,mul(f[i - 1],b[i + 1]));
	}
}
void CalcH() {
	memset(f,0,sizeof(f));
	memset(b,0,sizeof(b));
	memset(s,0,sizeof(s));
	int X = sum[2][M],Y,Z;
	for(int i = 0 ; i <= N ; ++i) {
		if(a[0][i + 1] == '0' && a[1][i + 1] == '0') continue;
		Y = sum[0][i],Z = sum[1][i];
		if(!X) {
			if(Y == 0 && Z == 0) f[i] = 1;
			else f[i] = 0;
		}
		else {
			f[i] = C(X + Y + Z - 1,X - 1);
		}
	}
	X = sum[3][M];
	for(int i = N + 1 ; i >= 1 ; --i) {
		if(a[0][i - 1] == '0' && a[1][i - 1] == '0') continue;
		Y = sum[0][N] - sum[0][i - 1],Z = sum[1][N] - sum[1][i - 1];
		if(!X) {
			if(Y == 0 && Z == 0) b[i] = 1;
			else b[i] = 0;
		}
		else {
			b[i] = C(X + Y + Z - 1,X - 1);
		}
	}
	for(int i = 1 ; i <= N ; ++i) s[i] = s[i - 1] + (a[0][i] == '1' && a[1][i] == '1');
	for(int i = 0 ; i <= N ; ++i) {
		f[i] = mul(f[i],ipw[s[i]]);
	}
	for(int i = N + 1 ; i >= 1 ; --i) {
		b[i] = inc(b[i + 1],mul(b[i],pw[s[i - 1]]));
	}
	for(int i = 1 ; i <= N ; i++) {
		update(ans,mul(f[i - 1],b[i + 1]));
	}
}
void Solve() {
	read(N);read(M);
	for(int i = 0 ; i <= 3 ; ++i) scanf("%s",a[i] + 1);
	for(int i = 0 ; i <= 3 ; ++i) {
		int T = (i <= 1 ? N : M);
		for(int j = 1 ; j <= T ; ++j) {
			sum[i][j] = sum[i][j - 1] + (a[i][j] == '1');
		}
	}
	int T = 2 * (N + M) + 10;
	fac[0] = 1;
	for(int i = 1 ; i <= T ; ++i) {
		fac[i] = mul(fac[i - 1],i);
	}
	invfac[T] = fpow(fac[T],MOD - 2);
	for(int i = T - 1 ; i >= 0 ; --i) {
		invfac[i] = mul(invfac[i + 1],i + 1);
	}
	pw[0] = 1;
	T = max(N,M);
	for(int i = 1 ; i <= T ; ++i) {
		pw[i] = mul(pw[i - 1],2);
	}
	ipw[0] = 1;
	for(int i = 1 ; i <= T ; ++i) {
		ipw[i] = mul(ipw[i - 1],(MOD + 1) / 2);
	}
	CalcV();
	CalcH();
	if(!sum[0][N] && !sum[1][N] && !sum[2][M] && !sum[3][M]) ans = 1;
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

F - Squeezing Slimes

对于一个单个的区间来说，我们可以用

如果\(a = 2^{k}\)能用k次合成

如果\(2^{k} < a < 2^{k + 1}\)，能用k+1次合成

我们记录一下每个区间最左的元素被选了几次，最右的区间选了几次

如果是\(a = 2^{k}\)，那么两边就选了k次

另一种有两个情况，左边k次右边k+1次，左边k+1次右边k次

每个相邻的地方可以减少\(min(r_{i},l_{i + 1})\)

用一个dp实现就行

#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 100005
#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;
int a[MAXN],cnt[MAXN],val[MAXN][2];
int64 dp[MAXN][2];
void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) {
		read(a[i]);
		int k = -1,x = a[i];
		while(x) {++k;x >>= 1;}
		cnt[i] = k;
	}
	for(int i = 1 ; i <= N ; ++i) {
		dp[i][0] = 1e16;dp[i][1] = 1e16;
		int s = cnt[i],t = cnt[i];
		if(a[i] != (1 << cnt[i])) ++t;
		val[i][0] = s;val[i][1] = t;
		for(int j = 0 ; j <= 1 ; ++j) {
			dp[i][0] = min(dp[i][0],dp[i - 1][j] + t - min(t,val[i - 1][j]));
			dp[i][1] = min(dp[i][1],dp[i - 1][j] + t - min(s,val[i - 1][j]));
		}
	}
	out(min(dp[N][0],dp[N][1]));enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}