【AtCoder】ARC066

ARC066

C - Lining Up

判断是否合法即可，合法是(2^{lfloor frac{N}{2} floor})

不合法就是0

#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 eps 1e-10
#define MAXN 200005
//#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 = 1000000007;
int N;
int cnt[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;
}
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);
    int a;
    for(int i = 1 ; i <= N ; ++i) {
	read(a);cnt[a]++;
    }
    if(N & 1) {
	if(cnt[0] != 1) {puts("0");return;}
    }
    for(int i = N - 1 ; i >= 1 ; i -= 2) {
	if(cnt[i] != 2) {
	    puts("0");return;
	}
    }
    out(fpow(2,N / 2));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Xor Sum

按位dp，每一位的两个数要么就是1 1，要么就是1 0（不分顺序），要么就是0 0

可以发现，这三种情况如果有一位上是不同的，那算出来的是不同的u和v

然后我们可以限制两个数的和每一位是多少来dp，因为和肯定大于异或和

设(dp[i][0 / 1][0/1])表示前i位，和有没有碰到上界，前面的和需不需要进位

#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 eps 1e-10
#define MAXN 200005
//#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 = 1000000007;
int64 N;
int dp[65][2][2];
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(N);
    dp[61][1][0] = 1;
    for(int i = 60 ; i >= 0 ; --i) {
	for(int j = 0 ; j < 2 ; ++j) {
	    for(int k = 0 ; k < 2 ; ++k) {
		int t = (N >> i) & 1,u;
		u = t;
		if(j == 0) u = 1;
		
		for(int h = 0 ; h <= u ; ++h) {
		    int p = j && (h == t);
		    if(k) {
			if(h == 0){
			    update(dp[i][p][1],dp[i + 1][j][k]);
			    update(dp[i][p][0],dp[i + 1][j][k]);
			}
			else {
			    update(dp[i][p][1],dp[i + 1][j][k]);
			}
		    }
		    else {
			if(h == 0) {
			    update(dp[i][p][0],dp[i + 1][j][k]);
			}
			else {
			    update(dp[i][p][1],dp[i + 1][j][k]);
			    update(dp[i][p][0],dp[i + 1][j][k]);
			}
		    }
		}
	    }
	}
    }
    int ans = 0;
    for(int j = 0 ; j < 2 ; ++j) {
	update(ans,dp[0][j][0]);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Addition and Subtraction Hard

发现括号嵌套超过三个总可以改成嵌套两个

然后我们就可以设(dp[i][j])表示第i位之前有j个没闭合的括号，有一个括号符号改变一次正反，我们只能在初始符号是减号的地方加括号，因为加号的后面加括号没什么用

#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 eps 1e-10
#define MAXN 100005
//#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;
int64 dp[MAXN][3];
int64 A[MAXN];
char op[MAXN][5];
void update(int64 &x,int64 y) {
    x = max(x,y);
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
	read(A[i]);
	if(i != N) scanf("%s",op[i] + 1);
    }
    dp[1][0] = A[1];
    dp[1][1] = -1e18;
    dp[1][2] = -1e18;
    for(int i = 2 ; i <= N ; ++i) {
	for(int j = 0 ; j < 3 ; ++j) dp[i][j] = -1e18;
	for(int j = 0 ; j < 3 ; ++j) {
	    int t = (op[i - 1][1] == '-');
	    int k = (t ^ j) & 1;
	    int64 d = A[i];
	    if(k) d = -A[i];
	    if(t && j < 2) update(dp[i][j + 1],dp[i - 1][j] + d);
	    for(int h = 0 ; h <= j ; ++h) update(dp[i][h],dp[i - 1][j] + d);
	}
    }
    out(dp[N][0]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - Contest with Drinks Hard

(之前的题解)

我写的斜率维护，放弃了我最擅长的叉积维护，然后发现叉积维护也不会爆long long哦……

一写斜率维护我的代码就会莫名变长而且难写……行吧

我们看这题

推了推式子，发现这是个斜率的式子，但是斜率单增还要求最大值？啥我又得二分凸包……好烦……

然后我们求一个pre[x]表示[1,x]的最大分数，和一个suf[x]表示[x,N]里的最大分数

然后对于一个点枚举一个包含它的区间，计算取值

显然超时

那就放在分治上，左端点在左区间，右端点在右区间，把最大值处理成前后缀max，两边都是斜率优化

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#define fi first
#define se second
#define pii pair<int,int>
//#define ivorysi
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define MAXN 300005
using namespace std;
typedef long long int64;
typedef double db;
typedef unsigned int u32;
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;
int64 T[MAXN],pre[MAXN],suf[MAXN],sum[MAXN],ans[MAXN],cal[MAXN];
struct Point {
	int64 x,y;
	Point(){}
	Point(int64 _x,int64 _y) {
		x = _x;y = _y;
	}
}que[MAXN],pre_pos[MAXN],suf_pos[MAXN];
struct qry_node {
	int id,pos;
	int64 v;
}qry[MAXN],tmp1[MAXN],tmp2[MAXN];
bool slope(Point a,Point b,Point c) {
	return (c.y - b.y) * (b.x - a.x) > (b.y - a.y) *  (c.x - b.x);
}
void DC(int l,int r,int ql,int qr) {
	if(qr < ql) return;
	if(l == r) return;
	int mid = (l + r) >> 1;
	int tot = 0;
	que[++tot] = pre_pos[l - 1];
	for(int i = l ; i <= mid ; ++i) {
		while(tot > 1) {
			if(slope(que[tot - 1],que[tot],pre_pos[i])) --tot;
			else break;
		}
		que[++tot] = pre_pos[i];
	}
	for(int i = r + 1 ; i >= mid + 2; --i) {
		int L = 1,R = tot;
		while(L < R) {
			int mid = (L + R) >> 1;
			if((que[mid + 1].y - que[mid].y) >= 1LL * i * (que[mid + 1].x - que[mid].x)) 
				L = mid + 1;
			else R = mid;
		}
		L = que[L].x;
		cal[i] = suf[i] + pre[L] - sum[i - 1] + sum[L] + 1LL * (i - L - 1) * (i - L) / 2;
	}
	tot = 0;
	que[++tot] = suf_pos[r + 1];
	for(int i = r ; i > mid ; --i) {
		while(tot > 1) {
			if(slope(suf_pos[i],que[tot],que[tot - 1])) --tot;
			else break;
		}
		que[++tot] = suf_pos[i];
	}
	for(int i = l - 1 ; i < mid ; ++i) {
		int L = 1,R = tot;
		while(L < R) {
			int mid = (L + R) >> 1;
			if((que[mid].y - que[mid + 1].y) >= 1LL * i * (que[mid].x - que[mid + 1].x))
				R = mid;
			else L = mid + 1;
		}
		L = que[L].x;
		cal[i] = pre[i] + suf[L] - sum[L - 1] + sum[i] + 1LL * (L - i - 1) * (L - i) / 2;
	}
	for(int i = l ; i < mid ; ++i) cal[i] = max(cal[i - 1],cal[i]);
	for(int i = r ; i > mid + 1 ; --i) cal[i] = max(cal[i + 1],cal[i]);
	int t1 = 0,t2 = 0;
	for(int i = ql ; i <= qr ; ++i) {
		if(qry[i].pos <= mid) {
			ans[qry[i].id] = max(ans[qry[i].id],cal[qry[i].pos - 1] + T[qry[i].pos] - qry[i].v);
			tmp1[++t1] = qry[i];
		}
		else {
			ans[qry[i].id] = max(ans[qry[i].id],cal[qry[i].pos + 1] + T[qry[i].pos] - qry[i].v);
			tmp2[++t2] = qry[i];
		}
	}
	int p = ql - 1;
	for(int i = 1 ; i <= t1 ; ++i) qry[++p] = tmp1[i];
	for(int i = 1 ; i <= t2 ; ++i) qry[++p] = tmp2[i];
	DC(l,mid,ql,ql + t1 - 1);
	DC(mid + 1,r,ql + t1,qr);
}	

void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) read(T[i]);
	for(int i = 1 ; i <= N ; ++i) sum[i] = sum[i - 1] + T[i];
	int tot = 0;
	que[++tot] = Point(0,0);
	for(int i = 1 ; i <= N ; ++i) {
		int l = 1,r = tot;
		while(l < r) {
			int mid = (l + r) >> 1;
			if((que[mid + 1].y - que[mid].y) >= 1LL * i * (que[mid + 1].x - que[mid].x)) 
				l = mid + 1;
			else r = mid;
		}
		l = que[l].x;
		pre[i] = pre[l] + 1LL * (i - l) * (i - l + 1) / 2 - sum[i] + sum[l];
		pre[i] = max(pre[i],pre[i - 1]);
		Point p = Point(i,pre[i] + sum[i] + (1LL * i * (i - 1)) / 2);
		while(tot > 1) {
			if(slope(que[tot - 1],que[tot],p)) --tot;
			else break;
		}
		que[++tot] = p;
	}
	tot = 0;
	que[++tot] = Point(N + 1,(1LL * (N + 1) * (N + 2) / 2) - sum[N]);
	for(int i = N ; i >= 1 ; --i) {
		int l = 1,r = tot;
		while(l < r) {
			int mid = (l + r) >> 1;
			if((que[mid].y - que[mid + 1].y) >= 1LL * i * (que[mid].x - que[mid + 1].x))
				r = mid;
			else l = mid + 1;
		}
		l = que[l].x;
		suf[i] = suf[l] + (1LL * (l - i + 1) * (l - i) / 2) - sum[l - 1] + sum[i - 1];
		suf[i] = max(suf[i],suf[i + 1]);
		Point p = Point(i,suf[i] - sum[i - 1] + (1LL * i * (i + 1) / 2));
		while(tot > 1) {
			if(slope(p,que[tot],que[tot - 1])) --tot;
			else break;
		}
		que[++tot] = p;
	}
	for(int i = 0 ; i <= N ; ++i) {
		pre_pos[i] = Point(i,pre[i] + sum[i] + (1LL * i * (i + 1) / 2));
	}
	for(int i = 1 ; i <= N + 1 ; ++i) {
		suf_pos[i] = Point(i,suf[i] - sum[i - 1] + (1LL * i * (i - 1) / 2));
	}
	read(M);
	int p;int64 v;
	for(int i = 1 ; i <= M ; ++i) {
		read(p);read(v);
		qry[i] = (qry_node){i,p,v};
		ans[i] = pre[p - 1] + suf[p + 1];
	}
	DC(1,N,1,M);
	for(int i = 1 ; i <= M ; ++i) {
		out(ans[i]);enter;
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
	return 0;
}