Codeforces Round #618 (Div. 2) 【A - E】

题目来源：https://codeforces.com/contest/1300
D题猜的，不过我觉得猜的挺有道理的233
发现E题真的不是很难（码量不大），但是就是想不到…

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A - Non-zero

多判断几下 应该不难

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define mp(x,y) make_pair(x,y)
#define r(x) read(x)
#define rrr(x,y,z) read(x);read(y);read(z)
#define FOR(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
typedef long long LL;
typedef pair<int,int> pt;
const int N=2e5+5;
const int M=2e3+5;
const int INF=0x7fffffff;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1);
LL n,m;
int f[N];
template<class T>
inline void read(T &x)
{
    char c; x=1;
    while((c=getchar())<'0'||c>'9') if(c=='-') x=-1;
    T res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=res*10+c-'0';
    x*=res;
}
int main()
{
    int t;
    r(t);
    while(t--){
        r(n);
        LL sum=0;
        int cnt=0;
        LL ans=0;
        FOR(i,1,n){
            r(f[i]);
            sum+=f[i];
            if(f[i]==0) cnt++;
        }
        if(cnt!=0){
            ans+=cnt;
            if(sum+cnt==0) ans++;
        }
        else if(sum==0){
            bool flag=1;
            FOR(i,1,n){
                if(f[i]!=-1) flag=0;
            }
            if(flag) ans+=2;
            else ans++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

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B - Assigning to Classes

就排个序 取中间两个值的差就好了

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define mp(x,y) make_pair(x,y)
#define r(x) read(x)
#define rrr(x,y,z) read(x);read(y);read(z)
#define FOR(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
typedef long long LL;
typedef pair<int,int> pt;
const int N=2e5+5;
const int M=2e3+5;
const int INF=0x7fffffff;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1);
LL n,m;
int f[N];
template<class T>
inline void read(T &x)
{
    char c; x=1;
    while((c=getchar())<'0'||c>'9') if(c=='-') x=-1;
    T res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=res*10+c-'0';
    x*=res;
}
int main()
{
    int t;
    r(t);
    while(t--){
        r(n);
        FOR(i,1,2*n) r(f[i]);
        sort(f+1,f+n*2+1);
        cout<<f[n+1]-f[n]<<endl;
    }
    return 0;
}

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C - Anu Has a Function

我太憨了，居然用线段树写，其实一个前缀or一个后缀or就够了。
就是很容易发现，其实这个最终值只和第一个的值有关，我们就枚举第一个是哪个数（假如是x），然后其他的数的or 假如是y ，那这种情况的结果就是 x-(x&y) 注意优先级 要打括号,怎么求除这个数之外的or呢？两个前缀和，别像我一样用线段树，杀鸡焉用牛刀…

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define mp(x,y) make_pair(x,y)
#define r(x) read(x)
#define rrr(x,y,z) read(x);read(y);read(z)
#define FOR(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
typedef long long LL;
typedef pair<int,int> pt;
const int N=2e5+5;
const int M=2e3+5;
const int INF=0x7fffffff;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1);
LL n,m;
int f[N];
int sum[N<<2];
template<class T>
inline void read(T &x)
{
    char c; x=1;
    while((c=getchar())<'0'||c>'9') if(c=='-') x=-1;
    T res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=res*10+c-'0';
    x*=res;
}
void build(int k,int l,int r)
{
    if(l==r){
        sum[k]=f[l];
        return ;
    }
    int mid=(l+r)>>1;
    build(ls); build(rs);
    sum[k]=sum[k<<1]|sum[k<<1|1];
}
int query(int k,int l,int r,int x,int y)
{
    if(x<=l&&r<=y){
        return sum[k];
    }
    int mid=(l+r)>>1;
    int res=0;
    if(mid>=x) res|=query(ls,x,y);
    if(mid<y) res|=query(rs,x,y);
    return res;
}
int main()
{
    int t;
    r(n);
    int pos=0;
    FOR(i,1,n){
        r(f[i]);
    }
    build(1,1,n);
    int maxx=-1;
    FOR(i,1,n){
        int now=0;
        if(i>1) now|=query(1,1,n,1,i-1);
        if(i<n) now|=query(1,1,n,i+1,n);
        //cout<<f[i]<<' '<<now<<endl;
        int res=f[i]-(f[i]&now);
        //cout<<res<<endl;
        if(maxx<res){
            pos=i;
            maxx=res;
        }
    }
    cout<<f[pos];
    FOR(i,1,n){
        if(i!=pos) cout<<' '<<f[i];
    }
    cout<<endl;
    return 0;
}

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D - Aerodynamic

这题题目看了40分钟…
猜的结论：要有偶数个点，其次对边要平行且相等
判断的话用向量判断最好不过了，少了精度的干扰。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define mp(x,y) make_pair(x,y)
#define r(x) read(x)
#define rrr(x,y,z) read(x);read(y);read(z)
#define FOR(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
typedef long long LL;
typedef pair<int,int> pt;
const int N=2e5+5;
const int M=2e3+5;
const int INF=0x7fffffff;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1);
LL n,m;
struct point
{
    LL x,y;
}f[N];
template<class T>
inline void read(T &x)
{
    char c; x=1;
    while((c=getchar())<'0'||c>'9') if(c=='-') x=-1;
    T res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=res*10+c-'0';
    x*=res;
}
double dis(point p1,point p2)
{

}
int main()
{
    r(n);
    FOR(i,1,n){
        r(f[i].x); r(f[i].y);
    }
    f[n+1]=f[1];
    bool flag=1;
    if(n&1){
        cout<<"NO\n";
        return 0;
    }
    FOR(i,1,n/2){
        point p1=f[i],p2=f[i+1],p3=f[i+n/2],p4=f[i+n/2+1];
        if(p2.x-p1.x==p3.x-p4.x&&p2.y-p1.y==p3.y-p4.y){
            flag=1;
        }
        else{
            flag=0;
            break;
        }
    }
    if(flag) cout<<"YES\n";
    else cout<<"NO\n";
    return 0;
}

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E - Water Balance

思路就是从左到右扫，每次将一个点加进去 和前面的比较 如果当前的点小，那就可以和前一个合并 让前一个变小，用栈实现都可以，我手动写栈是因为 stack不好输出233

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define mp(x,y) make_pair(x,y)
#define r(x) read(x)
#define rrr(x,y,z) read(x);read(y);read(z)
#define FOR(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
typedef long long LL;
typedef pair<double,int> pt;
const int N=1e6+5;
const int M=2e3+5;
const int INF=0x7fffffff;
const int mod=998244353;
const double eps=1e-8;
const double pi=acos(-1);
LL n,m;
int f[N];
pt ans[N];
template<class T>
inline void read(T &x)
{
    char c; x=1;
    while((c=getchar())<'0'||c>'9') if(c=='-') x=-1;
    T res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=res*10+c-'0';
    x*=res;
}
int main()
{
    r(n);
    FOR(i,1,n) r(f[i]);
    int cnt=0;
    ans[++cnt]=mp(f[1],1);
    FOR(i,2,n){
        if(f[i]<ans[cnt].first){
            double res=(ans[cnt].first*ans[cnt].second+f[i])*1.0/(ans[cnt].second+1);
            ans[cnt]=mp(res,ans[cnt].second+1);
            while(cnt>=2&&ans[cnt].first<ans[cnt-1].first){
                res=(ans[cnt].first*ans[cnt].second+ans[cnt-1].first*ans[cnt-1].second)*1.0/(ans[cnt].second+ans[cnt-1].second);
                ans[--cnt]=mp(res,ans[cnt].second+ans[cnt-1].second);
            }
        }
        else ans[++cnt]=mp(f[i],1);
    }
    FOR(i,1,cnt){
        FOR(j,1,ans[i].second) printf("%.9f\n",ans[i].first);
    }
    cout<<endl;
    return 0;
}

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