标签：iterator read stream cout lower 关系 fir 删掉 bitset
A：考虑每一位的改变情况，分为强制变为1、强制变为0、不变、反转四种，得到这个之后and一发or一发xor一发就行了。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 500010
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 a[20];
signed main()
{
int n=read(),x=0;
while (n--)
{
char c=getchar();
while (c!=‘|‘&&c!=‘^‘&&c!=‘&‘) c=getchar();
int u=read();
for (int i=0;i<10;i++)
if (u&(1<<i))
{
if (c==‘|‘) a[i]=1;
if (c==‘^‘)
{
if (a[i]==0) a[i]=3;
else if (a[i]==1) a[i]=2;
else if (a[i]==2) a[i]=1;
else if (a[i]==3) a[i]=0;
}
}
else
{
if (c==‘&‘) a[i]=2;
}
}
cout<<3<<endl;
for (int i=1;i<=3;i++)
{
int x=0;
for (int j=0;j<10;j++)
if (a[j]==i) x|=1<<j;
if (i==2) cout<<‘&‘<<‘ ‘<<(1023^x)<<endl;
else
{
if (i==1) cout<<‘|‘;else cout<<‘^‘;
cout<<‘ ‘<<x<<endl;
}
}
return 0;
//NOTICE LONG LONG!!!!!
}
B：先按k=1的情况处理一下。然后若考虑首尾相接是否会超过m个，若会则删掉，若恰好有k个则继续删。最后如果只剩下一种数特殊讨论。坑点比较多。
#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],stk[N],top;
ll ans;
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read(),m=read(),k=read();
for (int i=1;i<=n;i++) a[i]=read();
int u=0;int cnt=0;
for (int i=1;i<=n;i++)
{
if (!top) stk[++top]=a[i],cnt++;
else
{
if (stk[top]==a[i]) cnt++;
else cnt=1;
stk[++top]=a[i];
}
if (cnt==m)
{
top-=m;ans+=1ll*k*m;cnt=0;
for (int j=top;j>=0;j--) if (stk[j]!=stk[top]) {cnt=top-j;break;}
}
}
u=top;memcpy(b,stk,sizeof(b));
int nnn=n;
n=u;memcpy(a,b,sizeof(a));
bool flag=0;
for (int i=2;i<=n;i++) if (a[i]!=a[1]) {flag=1;break;}
if (!flag) {cout<<1ll*nnn*k%m;return 0;}
int l=1,r=n;
while (l<=r)
{
int x=l,y=r;
if (a[l]!=a[r]) break;
while (a[x+1]==a[l]) x++;
while (a[y-1]==a[r]) y--;
if (x>=y)
{
ans+=(1ll*k*(r-l+1)/m)*m;
if (1ll*k*(r-l+1)%m==0) ans+=n-(r-l+1);
break;
}
if (x-l+1+r-y+1>=m)
{
ans+=1ll*(k-1)*m;
if (x-l+1+r-y+1>m) break;
else if (m==(x-l+1+r-y+1)) l=x+1,r=y-1;
else break;
}
else break;
}
ans=1ll*nnn*k-ans;
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
C：考虑建图，a向b连边表示a可以打败b。这样缩点后度数为0的点（显然只会存在一个）中的所有选手都可能取胜。考虑动态维护。每个SCC记录大小和各项的minmax。注意到所有SCC每一项都构成偏序关系(maxi<mini+1)，于是用一个set，新加入一个点时找到前驱后继，看新点能否与其互相打败而合并成一个新的SCC即可。具体实现参考了这份代码https://codeforces.com/contest/878/submission/31775120，感觉过于优美。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
#define ll long long
#define N 50010
#define M 12
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;
struct data
{
int size,min[M],max[M];
bool operator <(const data&a) const
{
for (int i=1;i<=m;i++) if (max[i]>a.min[i]) return 0;
return 1;
}
void merge(const data&a)
{
size+=a.size;
for (int i=1;i<=m;i++)
{
if (a.min[i]<min[i]) min[i]=a.min[i];
if (a.max[i]>max[i]) max[i]=a.max[i];
}
}
};
multiset<data> q;
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read(),m=read();
for (int i=1;i<=n;i++)
{
data p;p.size=1;
for (int j=1;j<=m;j++) p.min[j]=p.max[j]=read();
multiset<data>::iterator first_beat_p=q.lower_bound(p),first_p_cannot_beat=q.upper_bound(p);
while (first_beat_p!=first_p_cannot_beat)
{
p.merge(*first_beat_p);
first_beat_p=q.erase(first_beat_p);
}
q.insert(p);multiset<data>::iterator tmp=q.end();tmp--;
printf("%d\n",(*tmp).size);
}
return 0;
//NOTICE LONG LONG!!!!!
}
D：考虑所有数值都为01的情况。这样一共只有2k种本质不同的特征，可以bitset记录每种特征最后的答案直接暴力过去。不为01的话容易想到二分答案之类，实际上使用完全相同的做法得到01时的答案，最后类似于二分答案的做即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;
#define ll long long
#define N 100010
#define K 12
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,k,a[K][N],b[N][K],cnt,u;
bitset<(1<<K)> S[N];
bool cmp(const int &x,const int &y)
{
return a[x][u]<a[y][u];
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("d.in","r",stdin);
freopen("d.out","w",stdout);
#endif
n=read(),k=read(),q=read();
for (int i=0;i<k;i++)
for (int j=1;j<=n;j++)
a[i][j]=read();
for (int i=0;i<k;i++)
{
for (int j=0;j<(1<<k);j++)
S[i][j]=(j&(1<<i))>0;
}
for (u=1;u<=n;u++)
{
for (int j=0;j<k;j++) b[u][j]=j;
sort(b[u],b[u]+k,cmp);
}
cnt=k-1;
for (int i=1;i<=q;i++)
{
int op=read(),x=read()-1,y=read();
if (op==1) S[++cnt]=S[x]|S[y-1];
if (op==2) S[++cnt]=S[x]&S[y-1];
if (op==3)
{
u=0;
for (int j=k-1;j>=0;j--)
{
u|=1<<b[y][j];
if (S[x][u]) {printf("%d\n",a[b[y][j]][y]);break;}
}
}
}
return 0;
//NOTICE LONG LONG!!!!!
}
标签：iterator read stream cout lower 关系 fir 删掉 bitset
原文地址：https://www.cnblogs.com/Gloid/p/10473806.html
