Hello 2019 自闭记

　　A：8min才过？？？

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define int long long
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;
}
signed main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#endif*/
    char a=getc(),b=getc();
    for (int i=;i<;i++)
    {
        char x=getc(),y=getc();
        if (a==x||b==y) {cout<<"YES";return ;}
    }
    cout<<"NO";
    return ;
    //NOTICE LONG LONG!!!!!
}

　　B：WA了一发？？？

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define int long long
#define N 20
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,a[N];
bool flag=;
void dfs(int k,int s)
{
    if (flag) return;
    if (k>n&&s==) {flag=;return;}
    if (k>n) return;
    dfs(k+,(s+a[k])%);
    dfs(k+,(s-a[k]+)%);
}
signed main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#endif*/
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    dfs(,);
    if (flag) cout<<"YES";else cout<<"NO";
    return ;
    //NOTICE LONG LONG!!!!!
}

　　C：多余匹配括号去掉后设右括号个数为x，左括号个数为y，则匹配的括号序列对要求y1+y2=0 x1=0 y1+x2=0，瞎贪贪就行了。RE了一发40min才过？？？

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define int long long
#define N 500010
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,ans;
char s[N];
vector<int> b[N<<];
struct data
{
    int x,y;
}a[N];
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
    const char LL[]="%I64d\n";
#endif
    n=read();
    for (int i=;i<=n;i++)
    {
        scanf("%s",s+);int m=strlen(s+);
        int cnt=;
        for (int j=;j<=m;j++)
        {
            if (s[j]=='(') cnt++;else cnt--;
            a[i].x=min(a[i].x,cnt);
        }
        a[i].y=cnt;
        b[a[i].y+N].push_back(a[i].x);
    }
    // y1+y2=0 y1+x2>=0
    for (int i=;i<(N<<);i++) sort(b[i].begin(),b[i].end());
    int cnt=;
    for (int i=;i<b[N].size();i++) if (b[N][i]>=) cnt++;
    ans=cnt/;
    for (int i=N+;i<(N<<);i++)
    {
        int head=,tail=b[i].size();tail--;
        while (head<b[N-(i-N)].size()&&tail>=&&b[i][tail]>=)
        {
            while (head<b[N-(i-N)].size()&&i-N+b[N-(i-N)][head]<) head++;
            if (head>=b[N-(i-N)].size()) break;
            tail--;head++;ans++;
        }
    }
    cout<<ans;
    return ;
    //NOTICE LONG LONG!!!!!
}

　　D：暴力dp，对每个质因子做即可。看错题了一发？？？

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 1000000007
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;
}
ll n,prime[];
int m,cnt[],a[],f[][],g[][],inv[],t,tot=,ans=;
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;
}
void dfs(int k,ll s,ll u)
{
    if (k>t) {ans=(ans+s%P*u)%P;return;}
    for (a[k]=;a[k]<=cnt[k];a[k]++)
    {
        dfs(k+,s,u*f[k][a[k]]%P);
        s*=prime[k];
    }
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    const char LL[]="%I64d\n";
#endif
    cin>>n>>m;
    for (int i=;1ll*i*i<=n;i++)
    if (n%i==)
    {
        t++;prime[t]=i;
        while (n%i==) cnt[t]++,n/=i;
    }
    if (n>) prime[++t]=n,cnt[t]=;
    for (int i=;i<=;i++) inv[i]=ksm(i,P-);
    for (int i=;i<=t;i++)
    {
        memset(g[],,sizeof(g[]));g[][cnt[i]]=;
        for (int j=;j<=m;j++)
        {
            g[j][cnt[i]+]=;
            for (int k=cnt[i];k>=;k--)
            g[j][k]=(g[j][k+]+1ll*g[j-][k]*inv[k+])%P;
        }
        for (int k=;k<=cnt[i];k++) f[i][k]=g[m][k];
    }
    //for (int i=1;i<=t;i++) tot=1ll*tot*C[cnt[i]+m][cnt[i]]%P;tot=inv(tot);
    dfs(,,);
    //cout<<1ll*ans*tot%P;
    cout<<ans;
    return ;
    //NOTICE LONG LONG!!!!!
}

　　F感觉非常神仙？？？过了很长时间才明白不是什么乱七八糟的线段树合并而是bitset啊？？？但还是不会啊？？？怎么这么多人过啊？？？叉人还被抢啊？？？自闭了啊？？？

　　result：rank 422 rating +12

　　F：显然只需要维护集合中每个数出现次数的奇偶性，考虑使用bitset。3操作非常难实现，可以考虑改为维护每个数的倍数的奇偶性，这样就相当于and了。然后考虑怎么以这种方式统计答案。使用莫比乌斯函数容斥即可。好思博啊怎么我不会啊？？？

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;
#define N 100010
#define V 7001
bitset<V> a[N],b[V];
int n,m,mobius[V];
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 main()
{
#ifndef ONLINE_JUDGE
    freopen("f.in","r",stdin);
    freopen("f.out","w",stdout);
#endif
    n=read(),m=read();
    for (int i=;i<V;i++)
        for (int j=i*i;j<V;j+=i*i)
        mobius[j]=;
    for (int i=;i<V;i++) mobius[i]^=;
    for (int i=;i<V;i++)
        for (int j=i;j<V;j+=i)
        b[i][j]=mobius[j/i];
    while (m--)
    {
        int op=read();
        if (op==)
        {
            int x=read(),y=read();a[x]=;
            for (int i=;i*i<=y;i++)
            if (y%i==)
            {
                a[x][i]=;
                if (i*i!=y) a[x][y/i]=;
            }
        }
        if (op==)
        {
            int z=read(),x=read(),y=read();
            a[z]=a[x]^a[y];
        }
        if (op==)
        {
            int z=read(),x=read(),y=read();
            a[z]=a[x]&a[y];
        }
        if (op==)
        {
            int x=read(),y=read();
            printf("%d",(a[x]&b[y]).count()&);
        }
    }
    return ;
}
//f[x]=Σcnt[d] d|x
//

　　G：先考虑k=1时怎么做。设f[i][1]为i作为斯坦纳树（暂且就这么叫）的根（该点不一定被选中）时所有方案边数之和，f[i][0]为上述方案数；g[i][1]为i作为斯坦纳树的根（但目前选中的点的斯坦纳树实际上并不将其包含，且选中点集不为空）时所有方案边数之和，g[i][0]为上述方案数。转移考虑f必须是选中根或至少选中两棵子树，g只选中一棵子树。f可以不考虑限制后再减掉g。这个沙雕dp已经想了我2h了。

　　然后考虑怎么求k次方。容易想到在dp中记录各次方，（事实上刚才的dp中就记录了0次方和1次方）具体转移由二项式定理即可得，注意细节。那么暴力复杂度是O(nk²)，NTT优化后可以做到O(nklogk)。复杂度看上去就很不靠谱，并且甚至题就没给NTT模数。（但还是有神仙用NTT过掉了）

#include<bits/stdc++.h>
using namespace std;
#define P 998244353 //当然原题的模数是1e9+7
#define N 100010
#define M 210
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,p[N],f[N][M],g[N][M],size[N],C[M][M],fac[M],inv[M],p2[N],tmp[M],a[<<],b[<<],r[<<],t,ans,inv3,inv512;
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 inc(int &x,int y){x+=y;if (x>=P) x-=P;}
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 INV(int a){return ksm(a,P-);}
void DFT(int *a,int n,int g)
{
    for (int i=;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=;i<=n;i<<=)
    {
        int wn=ksm(g,(P-)/i);
        for (int j=;j<n;j+=i)
        {
            int w=;
            for (int k=j;k<j+(i>>);k++,w=1ll*w*wn%P)
            {
                int x=a[k],y=1ll*w*a[k+(i>>)]%P;
                a[k]=(x+y)%P,a[k+(i>>)]=(x-y+P)%P;
            }
        }
    }
}
void mul(int *a,int *b)
{
    DFT(a,<<,),DFT(b,<<,);
    for (int i=;i<(<<);i++) a[i]=1ll*a[i]*b[i]%P;
    DFT(a,<<,inv3);
    for (int i=;i<(<<);i++) a[i]=1ll*a[i]*inv512%P;
}
void dfs(int k,int from)
{
    f[k][]=;size[k]=;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from)
    {
        dfs(edge[i].to,k);
        memset(a,,sizeof(a)),memset(b,,sizeof(b));
        for (int j=;j<=m;j++) a[j]=inv[j],b[j]=1ll*inv[j]*(f[edge[i].to][j]+g[edge[i].to][j])%P;
        mul(a,b);for (int j=;j<=m;j++) tmp[j]=1ll*a[j]*fac[j]%P;
        for (int j=;j<=m;j++) inc(g[k][j],tmp[j]);
        memset(a,,sizeof(a));memset(b,,sizeof(b));
        for (int j=;j<=m;j++) a[j]=1ll*inv[j]*f[k][j]%P,b[j]=1ll*inv[j]*tmp[j]%P;
        mul(a,b);for (int j=;j<=m;j++) tmp[j]=1ll*a[j]*fac[j]%P;
        for (int j=;j<=m;j++) inc(f[k][j],tmp[j]);
        f[k][]=p2[size[k]+=size[edge[i].to]];
    }
    for (int j=;j<=m;j++) inc(f[k][j],P-g[k][j]);
    f[k][]--;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("g.in","r",stdin);
    freopen("g.out","w",stdout);
#endif
    n=read(),m=read();inv3=INV(),inv512=INV();
    C[][]=;
    for (int i=;i<=m;i++)
    {
        C[i][]=C[i][i]=;
        for (int j=;j<i;j++)
        C[i][j]=(C[i-][j-]+C[i-][j])%P;
    }
    fac[]=;for (int i=;i<=m;i++) fac[i]=1ll*fac[i-]*i%P;
    inv[]=inv[]=;for (int i=;i<=m;i++) inv[i]=P-1ll*inv[P%i]*(P/i)%P;
    for (int i=;i<=m;i++) inv[i]=1ll*inv[i-]*inv[i]%P;
    for (int i=;i<(<<);i++) r[i]=(r[i>>]>>)|(i&)*(<<);
    for (int i=;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    p2[]=;for (int i=;i<=n;i++) p2[i]=(p2[i-]<<)%P;
    dfs(,);
    for (int i=;i<=n;i++) inc(ans,f[i][m]);
    cout<<ans;
    return ;
}

　　换一种思路。https://www.cnblogs.com/Gloid/p/9553291.html 我们想起来还做过这么个题。搬上同样的做法即可。然后我发现我不会转移了，就非常开心的弃掉了。

　　E：见官方题解及讨论区一楼补充。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 100010
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,a[N],b[N],f[N],q[N],cnt;
bool flag[N];
vector<int> ans[N],tmp[N];
int solve(int op)
{
    for (int i=;i<=n;i++) q[i]=;
    int tail=;tmp[].clear();
    for (int i=;i<=n;i++)
    {
        int l=,r=tail,ans=;
        while (l<=r)
        {
            int mid=l+r>>;
            if (a[i]>a[q[mid]]) ans=mid,l=mid+;
            else r=mid-;
        }
        f[i]=q[ans];ans++;if (ans>tail) {tail++;if (op) tmp[tail].clear();}
        q[ans]=i;if (op) tmp[ans].push_back(a[i]);
    }
    if (op) for (int i=;i<=tail;i++) ans[++cnt]=tmp[i];
    return tail;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("e.in","r",stdin);
    freopen("e.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        n=read();
        for (int i=;i<=n;i++) a[i]=read();
        cnt=;
        while (n)
        {
            int LIS=solve(),end_LIS=q[LIS];
            for (int i=;i<=n;i++) flag[i]=;
            if ((1ll*LIS*(LIS+)>>)>n)
            {
                cnt++;ans[cnt].clear();
                int x=end_LIS;
                while (x)
                {
                    ans[cnt].push_back(a[x]);
                    flag[x]=;x=f[x];
                }
                reverse(ans[cnt].begin(),ans[cnt].end());
                int m=;
                   for (int i=;i<=n;i++) if (!flag[i]) b[++m]=a[i];
                n=m;for (int i=;i<=n;i++) a[i]=b[i];
            }
            else {solve();break;}
        }
        printf("%d\n",cnt);
        for (int i=;i<=cnt;i++)
        {
            printf("%d ",ans[i].size());
            for (int j=;j<ans[i].size();j++)
            printf("%d ",ans[i][j]);
            printf("\n");
        }
    }
    return ;
}