【AtCoder】ARC072

时间 2019-12-09

标签 AtCoder arc072 arc 繁體版

原文原文链接

ARC072

C - Sequence

直接认为一个数是正的，或者第一个数是负的，每次将不合法的负数前缀和改为+1正数前缀和改为-1c++

#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 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 a[MAXN],s[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    int64 tmp = 0,ans = 1e18;
    for(int i = 1 ; i <= N ; ++i) {
    s[i] = s[i - 1] + a[i];
    if(i & 1) {
        if(s[i] >= 0) {tmp += s[i] + 1;s[i] = -1;}
    }
    else {
        if(s[i] <= 0) {tmp += 1 - s[i];s[i] = 1;}
    }
    }
    ans = min(ans,tmp);
    tmp = 0;
    for(int i = 1 ; i <= N ; ++i) {
    s[i] = s[i - 1] + a[i];
    if(i & 1) {
        if(s[i] <= 0) {tmp += 1 - s[i];s[i] = 1;}
    }
    else {
        if(s[i] >= 0) {tmp += s[i] + 1;s[i] = -1;}
    }
    }
    ans = min(ans,tmp);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Alice&Brown

作atc的博弈论就像挖金子，你挖到了以后才知道它埋得有多浅，而你挖到以前，都刨过好几个天坑了= =spa

必败策略就是\(X\)和\(Y\)相差不超过1！！！code

其他都必胜队列

#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 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);
}
int64 X,Y;
void Solve() {
    read(X);read(Y);
    if(abs(X - Y) <= 1) {
    puts("Brown");
    }
    else {
    puts("Alice");
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Alice in linear land

咱们按照序列递推一遍，获得每一个初始指令进行后这个位置的值，询问一个p，设某个位置的值是\(f[p]\)，咱们就找一个小于等于\(f[p - 1]\)的值，使得\(p + 1,N\)里剩下的数没法使这个数到达终点get

反着递推就很简单了，设\(b[i]\)为后i个位置保证\(1 - b[i]\)能够达到的最大\(b[i]\)，那么新加一个数\(a[i - 1]\)，咱们获得\([a[i - 1] - b[i],a[i - 1] + b[i] ]\)都是能够到达的，咱们只要知足\(b[i] + 1 >= a[i - 1] - b[i]\)，咱们就能够用\(a[i - 1] + b[i]\)来更新\(b[i - 1]\)了it

#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 eps 1e-10
#define MAXN 500005
//#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,D,Q;
int a[MAXN],b[MAXN],f[MAXN];
void Solve() {
    read(N);read(D);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    for(int i = N ; i >= 1 ; --i) {
    if(a[i] - b[i + 1] <= b[i + 1] + 1) b[i] = a[i] + b[i + 1];
    else b[i] = b[i + 1];
    }
    f[0] = D;
    for(int i = 1 ; i <= N ; ++i) {
    f[i] = min(abs(f[i - 1] - a[i]),f[i - 1]);
    }
    read(Q);
    int p;
    for(int i = 1 ; i <= Q ; ++i) {
    read(p);
    if(f[p - 1] > b[p + 1]) puts("YES");
    else puts("NO");
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Dam

把一个点转化为一个向量\((v_i,t_i \times v_i)\)class

而后就变成了每次在队列后加一个向量，不断删前面的序列使得向量的终点的x = Lim

而后重新加的点开始两两合并向量直到造成一个凸包db

#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 eps 1e-10
#define MAXN 500005
//#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 Point {
    double x,y;
    Point(double _x = 0,double _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 double operator * (const Point &a,const Point &b) {
        return a.x * b.y - a.y * b.x;
    }
}que[MAXN];
int N,ql,qr;
double L,sum;
void Solve() {
    read(N);scanf("%lf",&L);
    double t,v;
    for(int i = 1 ; i <= N ; ++i) {
        scanf("%lf%lf",&t,&v);
        que[++qr] = Point(v,t * v);
        sum += t * v;
        double dec = 0;
        if(i != 1) {
            while(1) {
                if(dec >= v) break;
                if(que[ql].x <= v - dec) {
                    dec += que[ql].x;
                    sum -= que[ql].y;
                    ++ql;
                }
                else {
                    sum -= que[ql].y;
                    que[ql].y -= que[ql].y / que[ql].x * (v - dec);
                    que[ql].x -= v - dec;
                    sum += que[ql].y;
                    break;
                }
            }
        }
        printf("%.7lf\n",sum / L);
        while(ql <= qr - 1) {
            if(que[qr] * que[qr - 1] >= -1e-8) {
                que[qr - 1] = que[qr] + que[qr - 1];
                --qr;
            }
            else break;
        }
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}