【AtCoder】ARC082

C - Together

用一个数组记一下一个数给它本身，左右贡献都是1，看看哪个数的总贡献最大

#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];
int cnt[MAXN];
void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) read(A[i]);
	for(int i = 1 ; i <= N ; ++i) {
		if(A[i]) cnt[A[i] - 1]++;
		cnt[A[i]]++;
		cnt[A[i] + 1]++;
	}
	int res = 0;
	for(int i = 0 ; i <= 100001 ; ++i) res = max(cnt[i],res);
	out(res);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

D - Derangement

就是从前往后，如果遇见一个\(A[i] = i\)和前或后换一下

#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,A[MAXN];

void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) read(A[i]);
	int cnt = 0;
	for(int i = 1 ; i <= N ; ++i) {
		if(A[i] == i) {
			if(i == N) swap(A[i],A[i - 1]);
			else swap(A[i],A[i + 1]);
			++cnt;
		}
	}
	out(cnt);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

E - ConvexScore

转化一下问题可以发现，就是给定一个点集，如果这个点集可以构成一个凸包包括所有点（就是所有点不共线），那么就有1的贡献

初始方案数是\(2^{N}\)，去掉只有一个点的，和一个点都不选的

然后枚举点对来枚举直线，然后算在这条线上的有几个点，需要选至少两个点，然后从总的方案数减掉，注意去重

#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);
}
const int MOD = 998244353;
int N,ans;
struct Point {
	int x,y;
	Point(int _x = 0,int _y = 0) {
		x = _x;y = _y;
	}
	friend Point operator + (const Point &a,const Point &b) {
		return Point(a.x + b.x,a.y + b.y);
	}
	friend Point operator - (const Point &a,const Point &b) {
		return Point(a.x - b.x,a.y - b.y);
	}
	friend int operator * (const Point &a,const Point &b) {
		return a.x * b.y - a.y * b.x;
	}
	friend int dot(const Point &a,const Point &b) {
		return a.x * b.x + a.y * b.y;
	}
}P[205];
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 Solve() {
	read(N);
	ans = fpow(2,N);ans = inc(ans,MOD - N - 1);
	for(int i = 1 ; i <= N ; ++i) {
		read(P[i].x);read(P[i].y);
	}
	for(int i = 1 ; i <= N ; ++i) {
		for(int j = 1 ; j <= N ; ++j) {
			if(i == j || P[i].x > P[j].x || (P[i].x == P[j].x && P[i].y > P[j].y)) continue;
			int cnt = 2;
			for(int k = 1 ; k <= N ; ++k) {
				if(k == i || k == j) continue;
				if((P[k] - P[i]) * (P[k] - P[j]) == 0) {
					if(P[k].x < P[j].x || (P[k].x == P[j].x && P[k].y < P[j].y)) goto fail;
					++cnt;
				}
			}
			ans = inc(ans,MOD - inc(fpow(2,cnt),MOD - cnt - 1));
			fail:;
		}
	}
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

F - Sandglass

这个维护的话，初始是从A有0克和A有X克开始算

如果到一个端点，我们可以算出来到这里的最小值是需要初始值小于等于MinX得到的，同理，也可以算出最大值是需要初始值大于等于MaxX得到

如果在MinX和MaxX之间，就是以每克+1

#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 K,Q;
int64 X;
vector<pair<int64,int64> > v;
void Solve() {
	read(X);read(K);
	int64 r;
	for(int i = 1 ; i <= K ; ++i) {
		read(r);
		v.pb(mp(r,-1));
	}
	int64 a,b;
	read(Q);
	for(int i = 1 ; i <= Q ; ++i) {
		read(a);read(b);
		v.pb(mp(a,b));
	}
	sort(v.begin(),v.end());
	int64 MinX = 0,MinY = 0,MaxX = X,MaxY = X;
	int64 pre = 0,k = -1;
	for(auto t : v) {
		MinY = MinY + (t.fi - pre) * k;
		if(MinY < 0) {
			MinX -= MinY;
			MinY = 0;
		}
		if(MinY > X) MinY = X;
		MaxY = MaxY + (t.fi - pre) * k;
		if(MaxY > X) {
			MaxX -= (MaxY - X);
			MaxY = X;
		}
		if(MaxY < 0) MaxY = 0;
		if(t.se == -1) k *= -1;
		else {
			if(t.se <= MinX) {out(MinY);enter;}
			else if(t.se >= MaxX) {out(MaxY);enter;}
			else {out(MinY + t.se - MinX);enter;}
		}
		pre = t.fi;
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}