cf Round 594

A.Warrior and Archer(思维)

战士一定会ban掉当前边缘的位置。而战士和射手就会选择剩下的最远的两点。
我们让剩下的最远的两点最近就达到了均衡。
于是我们枚举战士ban掉的边缘，ban的次数是一定的。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

int a[N];

int main ()
{
    int n;
    scanf("%d",&n);
    FOR(i,,n) scanf("%d",a+i);
    sort(a+,a+n+);
    int ans=INF;
    int k=n-(n-)/;
    FOR(i,,n+-k) ans=min(ans,a[i+k-]-a[i]);
    printf("%d\n",ans);
    return ;
}

B.Max and Bike(二分)

二分时间t，然后用控制精度check就行了。关键是这题的单调性不明显。

另外用fabs控制精度不行，浮点数误差。。。以后就控制二分次数吧。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

int n;
double s, f, r, v;
bool check(double t)
{
    double S=v*t, l=*pi*r;
    double ss=S-floor(S/l)*l;
    double T=S+fabs(sin(ss//r))**r;
    return T>=f-s;
}
int main ()
{
    scanf("%d%lf%lf",&n,&r,&v);
    while (n--) {
        scanf("%lf%lf",&s,&f);
        double l=, r=1e12;
        int tot=;
        while (tot--) {
            double mid=(l+r)/;
            if (check(mid)) r=mid;
            else l=mid;
        }
        printf("%.7lf\n",l);
    }
    return ;
}

C.Edo and Magnets(贪心)

求平面n个点至多减少k个点后，用一个矩形覆盖它们，求这个矩形面积的最小值。(k<=10)

显然把边界的点先减掉最好，于是我们枚举四个边界上的点减少多少个后更新答案就ok了。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

struct node{int x, y;}p[N];
bool cmp1(int a,int b){return p[a].x<p[b].x;}
bool cmp2(int a,int b){return p[a].x>p[b].x;}
bool cmp3(int a,int b){return p[a].y<p[b].y;}
bool cmp4(int a,int b){return p[a].y>p[b].y;}
int pos1[N], pos2[N], pos3[N], pos4[N], last[N];

int main()
{
    int n, k;
    scanf("%d%d",&n,&k);
    FO(i,,n) {
        int x1,y1,x2,y2;
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        p[i].x=(x1+x2); p[i].y=(y1+y2);
        pos1[i]=pos2[i]=pos3[i]=pos4[i]=i;
    }
    sort(pos1,pos1+n,cmp1); sort(pos2,pos2+n,cmp2);
    sort(pos3,pos3+n,cmp3); sort(pos4,pos4+n,cmp4);
    int now=;
    LL ans = 1LL<<;
    FOR(a,,k) FOR(b,,k) FOR(c,,k) FOR(d,,k) {
        now++;
        int cnt = ;
        FO(i,,a) if(last[pos1[i]]!=now) last[pos1[i]]=now,cnt++;
        FO(i,,b) if(last[pos2[i]]!=now) last[pos2[i]]=now,cnt++;
        FO(i,,c) if(last[pos3[i]]!=now) last[pos3[i]]=now,cnt++;
        FO(i,,d) if(last[pos4[i]]!=now) last[pos4[i]]=now,cnt++;
        if(cnt!=k) continue;
        LL Maxx=-1LL<<,Maxy=-1LL<<,Minx=1LL<<,Miny=1LL<<;
        FO(i,,n) {
            if(last[i]!=now) {
                Maxx=max(Maxx,p[i].x*1LL);
                Minx=min(Minx,p[i].x*1LL);
                Maxy=max(Maxy,p[i].y*1LL);
                Miny=min(Miny,p[i].y*1LL);
            }
        }
        LL x=Maxx-Minx, y=Maxy-Miny;
        x=max(x,2LL); y=max(y,2LL);
        ans=min(ans,x*y);
    }
    printf("%lld\n",ans/);
}

D.REQ(BIT+积性函数+逆元)

询问区间积[l,r]的欧拉函数。n,q<=2e5,ai<=1e6

因为欧拉函数是积性函数，我们化解一下式子，把不同的素因子拿出来维护一下就行了。
离散化后用树状数组离线处理区间[l,r]内的不同质因子，类似于HH的项链。
之后求答案的时候用逆元搞搞就行了。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

typedef struct{int l, r, id;}Node;
Node qq[N];
int a[N], pri[N*], n;
LL sum[N], isum[N], mul[N], imul[N], tree[N], res[N];
queue<int> to[N];
map<int, int> vis;
VI vc[N];

void get_prim()
{
    FOR(i,,) {
        if (!pri[i]) pri[++pri[]]=i, vis[i]=pri[];
        for (int j=; j<=pri[]&&pri[j]<=/i; ++j) {
            pri[pri[j]*i]=;
            if (i%pri[j]==) break;
        }
    }
}
bool comp(Node a, Node b){return a.l<b.l;}
LL inv(LL a, LL m){
    if (a==) return ;
    return inv(m%a,m)*(m-m/a)%m;
}
void add(int x, LL val)
{
    while (x<=n) tree[x]=tree[x]*val%MOD, x+=lowbit(x);
}
LL query(int x)
{
    LL ans=;
    while (x) ans=ans*tree[x]%MOD, x-=lowbit(x);
    return ans;
}
int main ()
{
    get_prim();
    FO(i,,pri[]) mul[i]=(LL)(pri[i]-)*inv(pri[i],MOD)%MOD, imul[i]=(LL)pri[i]*inv(pri[i]-,MOD)%MOD;
    int q;
    n=Scan();
    FOR(i,,n) tree[i]=;
    sum[]=isum[]=;
    FOR(i,,n) {
        a[i]=Scan(), sum[i]=sum[i-]*a[i]%MOD, isum[i]=isum[i-]*inv(a[i],MOD)%MOD;
        int temp=a[i];
        for (int j=; pri[j]*pri[j]<=temp; ++j) {
            if (temp%pri[j]==) {
                to[j].push(i), vc[i].pb(j), temp/=pri[j];
                while (temp%pri[j]==) temp/=pri[j];
            }
        }
        if (temp>) to[vis[temp]].push(i), vc[i].pb(vis[temp]);
    }
    q=Scan();
    FOR(i,,q) qq[i].l=Scan(), qq[i].r=Scan(), qq[i].id=i;
    sort(qq+,qq+q+,comp);
    FO(i,,pri[]) if (!to[i].empty()) add(to[i].front(),mul[i]);
    int now=;
    FOR(i,,q) {
        while (now<qq[i].l) {
            int size=vc[now].size();
            for (int j=; j<size; ++j) {
                int k=vc[now][j];
                add(to[k].front(),imul[k]), to[k].pop();
                if (!to[k].empty()) add(to[k].front(),mul[k]);
            }
            now++;
        }
        res[qq[i].id]=sum[qq[i].r]*isum[qq[i].l-]%MOD*query(qq[i].r)%MOD;
    }
    FOR(i,,q) printf("%lld\n",res[i]);
    return ;
}

E.Cutting the Line(待填坑)