Codeforces Round #670 (Div. 2)

A

#include "bits/stdc++.h"
using namespace std;
#define all(v) (v).begin(), (v).end()
#define io ios::sync_with_stdio(0)
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define rson rt << 1 | 1, mid + 1, r
#define lson rt << 1, l, mid
#define lll __int128
#define lowbit(i) ((-i) & (i))
#define pii pair<int, int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back


template<class T>void read(T &x)
{
    x=0;
    int f=0;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        f|=(ch=='-');
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    x=f?-x:x;
    return;
}

#define ull unsigned long long
#define eps 1e-12
#define sc(x) scanf("%lld", &(x))
#define mem(a,b) memset(a,b,sizeof(a))
#define endl "
"
#define inf 0x3f3f3f3f
#define ll long long
const int mod = 1e9+7;
const int maxn=5e5+10;


int a[maxn];

void work()
{
    mem(a,0);
    int n;
    cin>>n;
    rep(i,1,n)
    {
        int now;
        cin>>now;
        a[now]++;
    }
    int ans=0;
    rep(i,0,100)
    {
        if(a[i]==0)
        {
            ans+=i;
            break;
        }
        a[i]--;
    }
    rep(i,0,100)
    {
        if(a[i]==0) {ans+=i;break;}
    }
    cout<<ans<<endl;
}

signed main()
{
    int t;
    cin>>t;
    while(t--)
    work();
}

B

当ans为正数时,正数取前5大,负数取前5小,枚举取答案
负数时, 负数取前五大,取答案
有0时ans最小为0

#include "bits/stdc++.h"
using namespace std;
#define all(v) (v).begin(), (v).end()
#define io ios::sync_with_stdio(0)
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define rson rt << 1 | 1, mid + 1, r
#define lson rt << 1, l, mid
#define lll __int128
#define lowbit(i) ((-i) & (i))
#define pii pair<int, int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back


template<class T>void read(T &x)
{
    x=0;
    int f=0;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        f|=(ch=='-');
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    x=f?-x:x;
    return;
}

#define ull unsigned long long
#define eps 1e-12
#define sc(x) scanf("%lld", &(x))
#define mem(a,b) memset(a,b,sizeof(a))
#define endl "
"
#define inf 0x3f3f3f3f
#define ll long long
#define int long long
const int mod = 1e9+7;
const int maxn=5e5+10;


int a[maxn];
int b[maxn];
int c[maxn];

int ans;

void dfs(int dep,int st,int ed,int now)
{
//    cout<<st<<" "<<ed<<endl;
    if(dep==5)
    {
        ans=max(ans,now);
        return ;
    }
    rep(i,st,ed)
    {
        dfs(dep+1,i+1,ed,now*c[i]);
        dfs(dep,i+1,ed,now);
    }
    return ;
}

int cmp(int a,int b)
{
    return a>b;
}

void work()
{
    int la=0,lb=0,n;
    ans=-1e18;
    cin>>n;
    rep(i,1,n)
    {
        int now;
        cin>>now;
        if(now>0) a[++la]=now;
        else if(now<0) b[++lb]=now;
        else if(now==0) ans=0;
    }
    sort(a+1,a+1+la,cmp);
    sort(b+1,b+1+lb);
    int e1=min(la,1ll*5);
    int e2=min(lb,1ll*5);
    int l=0;
//    rep(i,1,e2) cout<<b[i]<<endl;
    rep(i,1,e1) c[++l]=a[i];
    rep(i,1,e2)
    {
        c[++l]=b[i];
//        cout<<b[i]<<" "<<c[l]<<endl;
    }
//    rep(i,1,l) cout<<c[i]<<endl;
    if(l>=5)
      dfs(0,1,l,1);

    sort(a+1,a+1+la,cmp);
    sort(b+1,b+1+lb,cmp);
    e1=min(la,1ll*5);
    e2=min(lb,1ll*5);
    l=0;
//    rep(i,1,e2) cout<<b[i]<<endl;
    rep(i,1,e1) c[++l]=a[i];
    rep(i,1,e2)
    {
        c[++l]=b[i];
//        cout<<b[i]<<" "<<c[l]<<endl;
    }
//    rep(i,1,l) cout<<c[i]<<endl;
    if(l>=5)
      dfs(0,1,l,1);

    cout<<ans<<endl;
}

signed main()
{
    int t;
    cin>>t;
    while(t--)
    work();
}

C

树的重心最多有 2 个且相邻
取一个重心上不是另一个重心相连的子树, 破坏掉连到另一个重心上即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
//ÇóÊ÷µÄÖØÐÄ
const int maxn=5e5+5;
struct node
{
    int to,nex;
} tr[maxn*4];
int n,ans;
int max_part[maxn];
int sizx[maxn],cnt;
int head[maxn];
int f[maxn];
void add(int u,int v)
{
    tr[cnt].to=v;
    tr[cnt].nex=head[u];
    head[u]=cnt++;
}
void dfs(int u,int fa)
{
    f[u]=fa;
    sizx[u]=1;
    max_part[u]=0;
    for(int i=head[u]; ~i; i=tr[i].nex)
    {
        int v=tr[i].to;
        if(v==fa)
        {
            continue;
        }
        dfs(v,u);
        sizx[u]+=sizx[v];
        max_part[u]=max(max_part[u],sizx[v]);
    }
    max_part[u]=max(max_part[u],n-sizx[u]);
    ans=min(max_part[u],ans);
}

int kk=0,ansv=0;
void dfs1(int u,int fa)
{
    if(kk) return ;
    for(int i=head[u]; ~i; i=tr[i].nex)
    {
        int v=tr[i].to;
        if(v==fa) continue;
        if(sizx[v]==1)
        {
            ansv=v;
            kk=1;
            return ;
        }
        dfs1(v,u);
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=1;i<=cnt;i++)
        {
            head[i]=-1;
            tr[i].nex=0,tr[i].to=0;

        }
        cnt=0;
        ans=0x3f3f3f3f3f;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            f[i]=0;
            max_part[i]=0;
            head[i]=-1;
            sizx[i]=0;
        }
        for(int i=1; i<n; i++)
        {
            int u,v;
            scanf("%d %d",&u,&v);
            add(u,v);
            add(v,u);
        }
        dfs(1,0);
        int root,minx=0x3f3f3f3f;
        int num=0;
        for(int i=1;i<=n;i++)
        {
            if(max_part[i]==ans)
            {
                root=i;
                break;
            }
        }

        int root2;
        for(int i=1;i<=n;i++)
        {
            if(max_part[i]==ans)
                root2=i;
        }

        int mx=0,pos;
        for(int i=head[root];~i;i=tr[i].nex)
        {
            int v=tr[i].to;
//            cout<<v<<endl;
            if(v!=root2)
            {
                pos=v;
                break;
            }
        }

        cout<<pos<<" "<<root<<"
";
        cout<<pos<<" "<<root2<<"
";
    }
}

D

维护差分
应为 b 是非递减, c 是非递增
所以答案应该在 bn 和 c1 中取得
把正差分加到 b 上, 负差分加到 c 上为最优
记正差分的和为sum, b1 为 x, c1 为 y
所以 x+y=a1 bn=x+sum
所以 ans=max( x+sum, a1-x )
且当x+sum=a1-x时取得最小值

#include "bits/stdc++.h"
using namespace std;
#define all(v) (v).begin(), (v).end()
#define io ios::sync_with_stdio(0)
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define rson rt << 1 | 1, mid + 1, r
#define lson rt << 1, l, mid
#define lll __int128
#define lowbit(i) ((-i) & (i))
#define pii pair<int, int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define int long long

template<class T>void read(T &x)
{
    x=0;
    int f=0;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        f|=(ch=='-');
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    x=f?-x:x;
    return;
}

#define ull unsigned long long
#define eps 1e-12
#define sc(x) scanf("%lld", &(x))
#define mem(a,b) memset(a,b,sizeof(a))
#define endl "
"
#define inf 0x3f3f3f3f
#define ll long long
#define dbug cout<<"here
";

const int mod = 1e9+7;
const int maxn=5e5+10;



int a[maxn];
int b[maxn];


void work()
{
    int n,q;
    read(n);
    rep(i,1,n) read(a[i]);
    int sum=0;
    rep(i,1,n) b[i]=a[i]-a[i-1];
    rep(i,1,n) if(b[i]>0&&i!=1) sum+=b[i];
    int x;
    x=(b[1]-sum)/2;
    int ans=max(x+sum,b[1]-x);
    printf("%lld
",ans);
    read(q);
    while(q--)
    {
        int l,r,d;
        read(l),read(r),read(d);
        if(l==1) b[1]+=d;
        else
        {
            if(b[l]>0&&b[l]+d<=0)
            {
                sum-=b[l];
                b[l]+=d;
            }
            else
            {
                if(b[l]>0)
                {
                    b[l]+=d;
                    sum+=d;
                }
                else
                {
                    b[l]+=d;
                    if(b[l]>0) sum+=b[l];
                }
            }
        }
        if(r!=n)
        {
            r++;
            d=-d;
            if(b[r]>0&&b[r]+d<=0)
            {
                sum-=b[r];
                b[r]+=d;
            }
            else
            {
                if(b[r]>0)
                {
                    b[r]+=d;
                    sum+=d;
                }
                else
                {
                    b[r]+=d;
                    if(b[r]>0) sum+=b[r];
                }
            }
        }
        x=(b[1]-sum)/2;
        ans=max(x+sum,b[1]-x);
        printf("%lld
",ans);
    }
}

signed main()
{

    work();
}

E

简单的想法就是枚举质数, 然后判断这个质数是否为x的质因子以及系数, 但是1E5内的质数接近1E4, 查询次数不允许.
所以对质数进行分块, 设m为质数的个数,每次删掉 sqrt(m) 的质数,同时维护一下应该剩余数的个数res, 每次删完 sqrt(m) 之后查询 1, 看和res是否相同, 确定最小的质因数, 之后同上枚举后面的每个质数