【AtCoder】ARC082

C - Together

用一個數組記一下一個數給它自己，左右貢獻都是1，看看哪一個數的總貢獻最大

#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

就是從前日後，若是碰見一個\(A[i] = i\)和前或後換一下

#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

轉化一下問題能夠發現，就是給定一個點集，若是這個點集能夠構成一個凸包包括全部點（就是全部點不共線），那麼就有1的貢獻
初始方案數是\(2^{N}\)，去掉只有一個點的，和一個點都不選的
而後枚舉點對來枚舉直線，而後算在這條線上的有幾個點，須要選至少兩個點，而後從總的方案數減掉，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

這個維護的話，初始是從A有0克和A有X克開始算
若是到一個端點，咱們能夠算出來到這裏的最小值是須要初始值小於等於MinX獲得的，同理，也能夠算出最大值是須要初始值大於等於MaxX獲得
若是在MinX和MaxX之間，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();
}