Educational Codeforces Round 56 Div. 2 翻车记

　　A：签到。

　　B：仅当只有一种字符时无法构成非回文串。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int T,n,cnt[];
char s[N];
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    T=read();
    while (T--)
    {
        scanf("%s",s+);n=strlen(s+);
        memset(cnt,,sizeof(cnt));
        for (int i=;i<=n;i++) cnt[s[i]-'a']++;
        int tot=;
        for (int i=;i<;i++) if (cnt[i]) tot++;
        if (tot==) cout<<-<<endl;
        else
        {
            for (int i=;i<;i++)
                for (int j=;j<=cnt[i];j++)
                putchar(i+'a');
            cout<<endl;
        }
    }
    return ;
}

　　C：在满足条件的前提下使每对数的差尽量大。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
ll read()
{
    ll x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n;
ll a[N],b[N],l,r;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();
    for (int i=;i<=(n>>);i++) a[i]=read();
    l=,r=;
    for (int i=;i<=(n>>);i++)
    {
        b[i]=l,b[n-i+]=a[i]-l;
        if (b[n-i+]>r) b[i]+=b[n-i+]-r,b[n-i+]=r;
        l=b[i],r=b[n-i+];
    }
    for (int i=;i<=n;i++) printf("%I64d ",b[i]);
    return ;
}

　　D：二分图染色，每个连通块贡献相乘即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 300010
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
ll read()
{
    ll x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int T,n,m,t,p[N],color[N],white,black;
bool flag;
struct data{int to,nxt;
}edge[N<<];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs(int k,int c)
{
    if (flag==) return;
    color[k]=c;if (c==) white++;else black++;
    for (int i=p[k];i;i=edge[i].nxt)
    {
        if (flag==) return;
        if (color[edge[i].to]==-) dfs(edge[i].to,c^);
        else if (color[edge[i].to]==c) {flag=;return;}
    }
}
int ksm(int a,int k)
{
    int s=;
    for (;k;k>>=,a=1ll*a*a%P)if (k&) s=1ll*s*a%P;
    return s;
}
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    T=read();
    while (T--)
    {
        n=read(),m=read();flag=;t=;
        for (int i=;i<=n;i++) p[i]=,color[i]=-;
        for (int i=;i<=m;i++)
        {
            int x=read(),y=read();
            addedge(x,y),addedge(y,x);
        }
        int cnt=,ans=;
        for (int i=;i<=n;i++)
        if (color[i]==-)
        {
            white=black=,cnt++,dfs(i,);
            ans=1ll*ans*(ksm(,white)+ksm(,black))%P;
        }
        if (flag==) ans=;
        printf("%d\n",ans);
    }
    return ;
}

　　这都啥思博题啊20min切完。E一眼树状数组套个权值线段树，犹豫了一下ECR的E会不会这么毒瘤，最后还是码了。30min之后疯狂RE22，因为空间根本开不下。这时候才想起来带修改之后树状数组套个treap和它复杂度是一样的，还能回收下空间，于是又开始码码码，过了20min终于A掉了。

　　E：将每个数在两个排列里的位置视为一个二维坐标。现在要做到就是插点/删点/矩形内点的个数。树状数组套treap即可，需要回收空间。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<ctime>
using namespace std;
#define ll long long
#define N 200010
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
ll read()
{
    ll x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,a[N],b[N],root[N],pos[N][],q[N<<],head,tail,t;
struct data{int ch[],p,x,s;
}tree[N<<];
int up(int k){tree[k].s=tree[lson].s+tree[rson].s+;}
void move(int &k,int p)
{
    int t=tree[k].ch[p];
    tree[k].ch[p]=tree[t].ch[!p],tree[t].ch[!p]=k,up(k),up(t),k=t;
}
void INS(int &k,int x)
{
    if (k==) {if (head<tail) k=q[++head];else k=++t;tree[k].ch[]=tree[k].ch[]=;tree[k].x=x;tree[k].p=rand();tree[k].s=;return;}
    tree[k].s++;
    if (tree[k].x<=x) {INS(rson,x);if (tree[rson].p>tree[k].p) move(k,);}
    else {INS(lson,x);if (tree[lson].p>tree[k].p) move(k,);}
}
void DEL(int &k,int x)
{
    if (tree[k].x==x)
    {
        if (lson==||rson==) {q[++tail]=k;k=lson|rson;return;}
        if (tree[lson].p>tree[rson].p) move(k,),DEL(rson,x);
        else move(k,),DEL(lson,x);
    }
    else if (tree[k].x<x) DEL(rson,x);
    else DEL(lson,x);
    up(k);
}
int find(int k,int x)
{
    if (!k) return ;
    if (tree[k].x<=x) return tree[lson].s++find(rson,x);
    else return find(lson,x);
}
void ins(int k,int x)
{
    while (k<=n) INS(root[k],x),k+=k&-k;
}
void del(int k,int x)
{
    while (k<=n) DEL(root[k],x),k+=k&-k;
}
int query(int k,int x)
{
    int ans=;
    while (k) ans+=find(root[k],x),k-=k&-k;
    return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    n=read(),m=read();
    srand(time());
    for (int i=;i<=n;i++) a[i]=read(),pos[a[i]][]=i;
    for (int i=;i<=n;i++) b[i]=read(),pos[b[i]][]=i;
    for (int i=;i<=n;i++) ins(i,pos[a[i]][]);
    for (int i=;i<=m;i++)
    {
        int op=read();
        if (op==)
        {
            int la=read(),ra=read(),lb=read(),rb=read();
            printf("%d\n",query(ra,rb)-query(ra,lb-)-query(la-,rb)+query(la-,lb-));
        }
        else
        {
            int x=read(),y=read();
            del(pos[b[x]][],x),del(pos[b[y]][],y),
            ins(pos[b[x]][],y),ins(pos[b[y]][],x);
            swap(b[x],b[y]);
        }
    }
    return ;
}

　　G：将两点间曼哈顿距离的绝对值展开讨论，可以发现对于每种情况，两点每维坐标的贡献与其原来该维坐标的关系都是相同的（即正或负），于是线段树暴力维护2^k种状态即可。我也不知道为什么两个树套树我只写了50min（bit套树根本就没什么码量吧），一个裸的线段树我能写40min。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<ctime>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,k,a[N][];
struct data{int l,r,f[],g[];
}tree[N<<];
void dfs(int node,int x,int p,int s,int sum)
{
    if (p==k) {tree[node].f[sum]=tree[node].g[sum]=s;return;}
    dfs(node,x,p+,s-a[x][p],sum);
    dfs(node,x,p+,s+a[x][p],sum|(<<p));
}
data merge(data x,data y)
{
    data u;u.l=x.l,u.r=y.r;
    for (int i=;i<(<<k);i++)
    u.f[i]=max(x.f[i],y.f[i]),u.g[i]=min(x.g[i],y.g[i]);
    return u;
}
void build(int k,int l,int r)
{
    tree[k].l=l,tree[k].r=r;
    if (l==r) {dfs(k,l,,,);return;}
    int mid=l+r>>;
    build(k<<,l,mid);
    build(k<<|,mid+,r);
    tree[k]=merge(tree[k<<],tree[k<<|]);
}
void modify(int k,int x)
{
    if (tree[k].l==tree[k].r) {dfs(k,x,,,);return;}
    int mid=tree[k].l+tree[k].r>>;
    if (x<=mid) modify(k<<,x);
    else modify(k<<|,x);
    tree[k]=merge(tree[k<<],tree[k<<|]);
}
data query(int k,int l,int r)
{
    if (tree[k].l==l&&tree[k].r==r) return tree[k];
    int mid=tree[k].l+tree[k].r>>;
    if (r<=mid) return query(k<<,l,r);
    else if (l>mid) return query(k<<|,l,r);
    else return merge(query(k<<,l,mid),query(k<<|,mid+,r));
}
int main()
{
    n=read(),k=read();
    for (int i=;i<=n;i++)
        for (int j=;j<k;j++)
        a[i][j]=read();
    build(,,n);
    m=read();
    for (int i=;i<=m;i++)
    {
        int op=read();
        if (op==)
        {
            int x=read();
            for (int j=;j<k;j++) a[x][j]=read();
            modify(,x);
        }
        else
        {
            int l=read(),r=read();
            data x=query(,l,r);
            int ans=;
            for (int i=;i<(<<k);i++)
            ans=max(ans,x.f[i]-x.g[i]);
            printf("%d\n",ans);
        }
    }
}

　　F：没看。

　　突然小号也变得橘里橘气。result：rank 10 rating +170