A：签到。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int b,k,a[100010];
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
b=read(),k=read();b&=1;
for (int i=1;i<=k;i++) a[i]=read();
if (b==0)
{
if (a[k]&1) cout<<"odd";
else cout<<"even";
}
else
{
int s=0;
for (int i=1;i<=k;i++) s+=a[i];
if (s&1) cout<<"odd";
else cout<<"even";
}
return 0;
//NOTICE LONG LONG!!!!!
}
B：显然先连接距离较小的点。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,k,a[N],b[N],ans;
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
n=read(),m=read(),k=read();
for (int i=1;i<=n;i++) a[i]=read();
ans=a[n]-a[1]+1;
for (int i=2;i<=n;i++) b[i]=a[i]-a[i-1];
sort(b+2,b+n+1);reverse(b+2,b+n+1);
for (int i=2;i<=k;i++) ans-=b[i]-1;
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
C：当a!=2k-1时，令k为满足2k-1>a的最小正整数，显然可以令b=2k-1^a使答案成为2k-1，并且显然不可能更优。a=2k-1时直接暴力打表。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int q,ans[1<<25];
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
q=read();
for (int i=1;i<=22;i++)
{
for (int j=1;j<(1<<i)-1;j++)
ans[(1<<i)-1]=max(ans[(1<<i)-1],gcd((1<<i)-1&j,(1<<i)-1^j));
}
ans[(1<<23)-1]=178481;
ans[(1<<24)-1]=5592405;
ans[(1<<25)-1]=1082401;
while (q--)
{
int x=read();
if (ans[x]>0) printf("%d
",ans[x]);
else
{
while (x!=(x&-x)) x^=x&-x;
printf("%d
",(x<<1)-1);
}
}
return 0;
//NOTICE LONG LONG!!!!!
}
E：https://www.cnblogs.com/Gloid/p/10060025.html 作为一个做过所谓原题的选手看了1h这个题感到十分自闭。事实上第一眼就想到了这个原题，发现并不一样之后就去梦游了。毕竟当时也不是用前缀和的做法做的，而是自己想了半天搞了个做法还觉得挺nb，反应不过来也挺正常。但现在感觉我这做法好像也能类似的搬过来？自闭了啊？
比较差分集合是否相同即可，因为每次操作相当于交换差分数组中相邻两数。注意特判首尾。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,q[N<<4],head,tail,cnt;
ll a[N],b[N];
bool flag[N];
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
n=read();
for (int i=1;i<=n;i++) a[i]=read();
for (int i=1;i<=n;i++) b[i]=read();
if (a[1]!=b[1]||a[n]!=b[n]) {cout<<"No";return 0;}
for (int i=n;i>=1;i--) a[i]-=a[i-1];
for (int i=n;i>=1;i--) b[i]-=b[i-1];
sort(a+1,a+n+1);
sort(b+1,b+n+1);
for (int i=1;i<=n;i++) if (a[i]!=b[i]) {cout<<"No";return 0;}
cout<<"Yes";
return 0;
//NOTICE LONG LONG!!!!!
}
D：注意到一定存在最优方案使得同种顺子最多出两次。然后就是普及组dp了。我这种弱智怎么可能注意的到啊。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,u,a[N],f[N][3][3];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif
n=read(),m=read();
for (int i=1;i<=n;i++) a[read()]++;
memset(f,200,sizeof(f));
f[0][0][0]=0;
for (int i=1;i<=m+2;i++)
for (int j=0;j<=2;j++)
for (int k=0;k<=2;k++)
for (int x=0;x<=2;x++)
if (k<=a[i]&&j+k<=a[i-1]&&j+k+x<=a[i-2])
f[i][j][k]=max(f[i][j][k],f[i-1][x][j]+x+(a[i-2]-j-k-x)/3);
cout<<f[m+2][0][0];
return 0;
//NOTICE LONG LONG!!!!!
}
F：对dfs序建线段树维护当前点到每个点的距离（非叶子设为inf），询问按dfs序离线，然后整棵树dfs一遍并维护距离即可，这个维护仅仅是线段树上的区间加和区间减。我觉得比D简单多了！算了云选手没资格说话。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 500010
#define inf 100000000000000000ll
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,fa[N],p[N],len[N],dfn[N],id[N],L[N<<2],R[N<<2],size[N],t,cnt,cur;
ll deep[N],ans[N],tree[N<<2],lazy[N<<2];
struct data
{
int to,len;
bool operator <(const data&a) const
{
return to<a.to;
}
};
vector<data> edge[N];
struct data2
{
int x,l,r,i;
bool operator <(const data2&a) const
{
return dfn[x]<dfn[a.x];
}
}q[N];
void addedge(int x,int y,int z){edge[x].push_back((data){y,z});}
void dfs(int k)
{
dfn[k]=++cnt;id[cnt]=k;size[k]=1;
for (int i=0;i<edge[k].size();i++)
{
deep[edge[k][i].to]=deep[k]+edge[k][i].len;
dfs(edge[k][i].to);
size[k]+=size[edge[k][i].to];
}
if (edge[k].size()) deep[k]=inf;
}
void up(int k){tree[k]=min(tree[k<<1],tree[k<<1|1]);}
void build(int k,int l,int r)
{
L[k]=l,R[k]=r;
if (l==r)
{
tree[k]=deep[id[l]];
return;
}
int mid=l+r>>1;
build(k<<1,l,mid);
build(k<<1|1,mid+1,r);
up(k);
}
void update(int k,ll x){tree[k]+=x;lazy[k]+=x;}
void down(int k){update(k<<1,lazy[k]),update(k<<1|1,lazy[k]),lazy[k]=0;}
void add(int k,int l,int r,int x)
{
if (l>r) return;
if (L[k]==l&&R[k]==r) {update(k,x);return;}
if (lazy[k]) down(k);
int mid=L[k]+R[k]>>1;
if (r<=mid) add(k<<1,l,r,x);
else if (l>mid) add(k<<1|1,l,r,x);
else add(k<<1,l,mid,x),add(k<<1|1,mid+1,r,x);
up(k);
}
ll query(int k,int l,int r)
{
if (L[k]==l&&R[k]==r) return tree[k];
if (lazy[k]) down(k);
int mid=L[k]+R[k]>>1;
if (r<=mid) return query(k<<1,l,r);
else if (l>mid) return query(k<<1|1,l,r);
else return min(query(k<<1,l,mid),query(k<<1|1,mid+1,r));
}
void work(int k)
{
while (k==q[cur+1].x) cur++,ans[q[cur].i]=query(1,q[cur].l,q[cur].r);
for (int i=0;i<edge[k].size();i++)
{
int to=edge[k][i].to;
add(1,dfn[to],dfn[to]+size[to]-1,-edge[k][i].len);
add(1,1,dfn[to]-1,edge[k][i].len);
add(1,dfn[to]+size[to],n,edge[k][i].len);
work(to);
add(1,dfn[to],dfn[to]+size[to]-1,edge[k][i].len);
add(1,1,dfn[to]-1,-edge[k][i].len);
add(1,dfn[to]+size[to],n,-edge[k][i].len);
}
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read(),m=read();
for (int i=2;i<=n;i++)
{
fa[i]=read();len[i]=read();
addedge(fa[i],i,len[i]);
}
for (int i=1;i<=n;i++) sort(edge[i].begin(),edge[i].end());
dfs(1);
for (int i=1;i<=m;i++) q[i].x=read(),q[i].l=read(),q[i].r=read(),q[i].i=i;
sort(q+1,q+m+1);
build(1,1,n);
work(1);
for (int i=1;i<=m;i++) printf("%I64d
",ans[i]);
return 0;
//NOTICE LONG LONG!!!!!
}
div1+div2的场从来没进过前400，自闭了。
result：rank 420 rating +22