【AtCoder】CODE FESTIVAL 2017 qual B

最近不知道为啥被安利了饥荒，但是不能再玩物丧志了，不能颓了
饥荒真好玩

A - XXFESTIVAL

CCFESTIVAL

#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('
')
#define MAXN 200005
#define eps 1e-10
//#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);
}
char s[55];
int L;
void Solve() {
	scanf("%s",s + 1);
	L = strlen(s + 1);
	for(int i = 1 ; i <= L - 8 ; ++i) {
		putchar(s[i]);
	}
	enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

B - Problem Set

当然是选择出毒瘤题了

#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('
')
#define MAXN 200005
#define eps 1e-10
//#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;
map<int,int> zz;
void Solve() {
	read(N);int a;
	for(int i = 1 ; i <= N ; ++i) {
		read(a);zz[a]++;
	}
	read(M);
	for(int i = 1 ; i <= M ; ++i) {
		read(a);
		if(zz[a]) {
			zz[a]--;
		}
		else {puts("NO");return;}
	}
	puts("YES");
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

C - 3 Steps

如果图是二分图，那么可以补的边是把点黑白染色后，所有黑点都可以跟白点连边

如果不是二分图，因为一个奇环，那么任意两点之间都可以连边，用最后图的边数减去原来的边数即可

#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('
')
#define MAXN 200005
#define eps 1e-10
//#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);
}
struct node {
	int to,next;
}E[MAXN];
int deg[MAXN],head[MAXN],sumE,N,M;
int cnt[2],col[MAXN];
void add(int u,int v) {
	E[++sumE].to = v;
	E[sumE].next = head[u];
	head[u] = sumE;
}
bool dfs(int u) {
	if(col[u] == -1) col[u] = 0;
	++cnt[col[u]];
	for(int i = head[u] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(col[v] == -1) {
			col[v] = col[u] ^ 1;
			if(!dfs(v)) return false;
		}
		else if(col[v] == col[u]) return false;
	}
	return true;
}
void Solve() {
	read(N);read(M);
	int a,b;
	for(int i = 1 ; i <= M ; ++i) {
		read(a);read(b);
		deg[a]++;deg[b]++;
		add(a,b);add(b,a);
	}
	memset(col,-1,sizeof(col));
	if(!dfs(1)) {
		out(1LL * N * (N - 1) / 2 - M);enter;
	}
	else {
		int64 ans = 0;
		for(int i = 1 ; i <= N ; ++i) {
			if(col[i]) ans += cnt[col[i] ^ 1] - deg[i];
		}
		out(ans);enter;
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

D - 101 to 010

可以用一个dp实现

划分1011111111（k个1）或11111111101（k个1），价值都是k - 1

如果这个位置连着的连续的1超过1个，最多就一种划分

如果这个位置连着的一个1，那么可以更新01的0前面一段的1

每个位置最多换两次，复杂度(O(N))

#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('
')
#define MAXN 500005
#define eps 1e-10
//#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,dp[MAXN],pre[MAXN];
char s[MAXN];
void Solve() {
	read(N);
	scanf("%s",s + 1);
	for(int i = 1 ; i <= N ; ++i) {
		if(s[i] == '0') pre[i] = i;
		else pre[i] = pre[i - 1];
	}
	for(int i = 1 ; i <= N ; ++i) {
		dp[i] = dp[i - 1];
		if(s[i] == '1') {
			if(pre[i] == 0) continue;
			int a = pre[i],b = pre[a - 1];
			if(b == a - 1) continue;
			dp[i] = max(dp[a - 2] + i - a,dp[i]);
			if(i - a == 1) {
				for(int j = b ; j <= a - 2 ; ++j) {
					dp[i] = max(dp[j] + a - j - 1,dp[i]);
				}
			}
		}
	}
	out(dp[N]);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

E - Popping Balls

一道有趣的计数题

就是还是改成二维平面上的点((A,B))求走到((0,0))的路径

相当于从((t - 1,0))画一条弧度为(frac{3pi}{4})的线和一条垂直于x轴的直线

这两条直线上方的部分可以往下走

((s - 1,0))同理

然后就相当于，枚举一个t，从起点走到边界时第一步是向下（因为如果进入了区域内还左走可以把这个区域往左平移，能走到的地方更多

然后在斜的边界上找一个点，枚举一个点(p,q)然后往左走走到s的边界同理，方案数是一个组合数的前缀和

#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('
')
#define MAXN 4005
#define eps 1e-10
//#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 MOD = 1000000007;
int A,B;
int C[MAXN][MAXN],sum[MAXN][MAXN],s[MAXN][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 update(int &x,int y) {
	x = inc(x,y);
}
void Solve() {
	read(A);read(B);
	C[0][0] = 1;
	sum[0][0] = 1;
	for(int i = 1 ; i <= A + B ; ++i) {
		C[i][0] = 1;
		sum[i][0] = 1;
		for(int j = 1 ; j <= i ; ++j) {
			C[i][j] = inc(C[i - 1][j],C[i - 1][j - 1]);
			sum[i][j] = inc(C[i][j],sum[i][j - 1]);
		}
	}
	for(int j = 0 ; j <= B ; ++j) {
		s[j][0] = 1;
		for(int i = 1 ; i <= A + B ; ++i) {
			s[j][i] = inc(s[j][i - 1],sum[j][min(i,j)]);
		}
	}
	int ans = 0;
	for(int t = A; t >= 0 ; --t) {
		for(int i = 0 ; i <= t ; ++i) {
			int w = C[B - 1][i];
			if(!i) update(ans,w);
			else {
				update(ans,mul(w,s[i - 1][t - i]));
			}
		}
	}
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

F - Largest Smallest Cyclic Shift

每次找最小的字符和最大的字符组合成一个新的字符串，把这个字符串当成一个新字符插入集合中

直到只剩一种字符

就是每次最小的字符串一定是循环串的开头，然后从后往前把这个字符串加上一个当前最大的字符串，最大的字符串就越靠前

#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('
')
#define MAXN 4005
#define eps 1e-10
//#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);
}
struct node {
	string s;int num;
	friend bool operator < (const node &a,const node &b) {
		return a.s < b.s;
	}
};
set<node> S;
int x,y,z;
void Solve() {
	read(x);read(y);read(z);
	if(x) S.insert((node){"a",x});
	if(y) S.insert((node){"b",y});
	if(z) S.insert((node){"c",z});
	while(1) {
		if(S.size() == 1) {
			node p = *S.begin();
			for(int i = 1 ; i <= p.num ; ++i) {
				cout << p.s;
			}
			enter;
			break;
		}
		node s0 = *S.begin(),t0 = *(--S.end());
		S.erase(S.begin());S.erase(--S.end());
		int k = min(s0.num,t0.num);
		S.insert((node){s0.s + t0.s,k});
		if(s0.num > k) S.insert((node){s0.s,s0.num - k});
		if(t0.num > k) S.insert((node){t0.s,t0.num - k});
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}