Codeforces Global Round 1 自闭记
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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
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