A.Duff and Weight Lifting(思维)
显然题目中只有一种情况可以合并 2^a+2^a=2^(a+1)。
我们把给出的mi排序一下，模拟合并操作即可。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=1000005; //Code begin... int a[N]; int main () { int n; scanf("%d",&n); FOR(i,1,n) scanf("%d",a+i); sort(a+1,a+n+1); int now=a[1], cnt=1, ans=0; FOR(i,2,n) { if (now==a[i]) cnt++; else { while (now!=a[i]&&cnt>1) { ++now; ans+=(cnt&1); cnt/=2; } if (now==a[i]) cnt++; else ans+=cnt, cnt=1, now=a[i]; } } while (cnt) ans+=(cnt&1), cnt/=2; printf("%d ",ans); return 0; }
B.Duff in Beach(DP)
这是一个计数问题。考虑DP。
考虑L<=n*k.
由于n*k<=1e6.我们把a数组变成b数组。令dp[i]表示以b[i]结尾的方法数。
那么dp[i]=sigma(上一段的数小于b[i]的dp[i])。
sigma我们可以用类似前缀和的办法维护一下。
考虑L>n*k.
此时数组太大不好直接构造，我们考虑把n*k的元素的最后n个数字平移到n*k后面的数字。
发现方法数是一样的。我们再统计每个数字可以平移多少次就OK了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=1000005; //Code begin... int to[N]; LL dp[N], sum[N]; PII a[N]; int comp(const void * a, const void * b){return *(int *)a-*(int *)b;} int main () { int n, k, cnt=0; LL l, ans=0; scanf("%d%lld%d",&n,&l,&k); FO(i,0,n) a[i].first=Scan(), a[i].second=i; sort(a,a+n); to[a[0].second]=++cnt; FO(i,1,n) { if (a[i].first==a[i-1].first) to[a[i].second]=cnt; else to[a[i].second]=++cnt; } int m=(l%n?l%n:n); FO(i,0,min((LL)n*k,l)) { if (i && i%n==0) { FOR(j,1,cnt) sum[j]=0; FO(j,i-n,i) sum[to[j%n]]=(sum[to[j%n]]+dp[j])%MOD; FOR(j,1,cnt) sum[j]=(sum[j-1]+sum[j])%MOD; } if (i<n) dp[i]=1; else dp[i]=(1+sum[to[i%n]])%MOD; ans=(ans+dp[i])%MOD; if (l>n*k && i>=n*k-n) ans=(ans+((l-i-1)/n%MOD)*dp[i]%MOD)%MOD; } printf("%lld ",ans); return 0; }
C.Duff in the Army(LCA)
题目要求维护树的路径上的标号最小的a个数。(a<=10)
维护路径上的某些东西一般有LCT,LCA,树形DP,树链刨分,树分治。
由于题目允许离线，于是我们可以类似LCA预处理出一坨东西。
fa[x][i][]里面的是x节点到x节点的2^i个父亲中的标号最小的10个数。
合并的时候归并一下，最后求答案的时候就类似求LCA一样边爬边归并。
细节很多。
复杂度O((n+q)*logn).
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=100005; //Code begin... struct Edge{int p, next;}edge[N<<1]; int head[N], cnt=1; int bin[20], fa[N][20][12], dep[N], temp[12], ans[12]; VI node[N]; void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;} void bin_init(){bin[0]=1; FO(i,1,20) bin[i]=bin[i-1]<<1;} void merge(int x, int nn) { int y=fa[x][nn-1][0], i=1, j=1, k=1; while (k<=10) { if (fa[x][nn-1][i]==0 && fa[y][nn-1][j]==0) break; if (fa[x][nn-1][i]==0) fa[x][nn][k++]=fa[y][nn-1][j++]; else if (fa[y][nn-1][j]==0) fa[x][nn][k++]=fa[x][nn-1][i++]; else { fa[x][nn][k++]=min(fa[x][nn-1][i], fa[y][nn-1][j]); if (fa[x][nn-1][i]<fa[y][nn-1][j]) ++i; else ++j; } } } void merge_(int x, int nn) { int i=1, j=1, k=1; mem(temp,0); FOR(l,1,10) temp[l]=ans[l]; mem(ans,0); while (k<=10) { if (fa[x][nn][i]==0&&temp[j]==0) break; if (fa[x][nn][i]==0) ans[k++]=temp[j++]; else if (temp[j]==0) ans[k++]=fa[x][nn][i++]; else { ans[k++]=min(fa[x][nn][i],temp[j]); if (fa[x][nn][i]<temp[j]) ++i; else if (fa[x][nn][i]>temp[j]) ++j; else ++i, ++j; } } } void dfs(int x, int fat) { fa[x][0][0]=fat; if (node[fat].size()) for (int i=0; i<min((int)node[fat].size(),10); ++i) fa[x][0][i+1]=node[fat][i]; for (int i=1; bin[i]<=dep[x]; ++i) fa[x][i][0]=fa[fa[x][i-1][0]][i-1][0], merge(x,i); for (int i=head[x]; i; i=edge[i].next) { int v=edge[i].p; if (v==fat) continue; dep[v]=dep[x]+1; dfs(v,x); } } void sol(int x, int y) { int l=0, j=0, k=1; while (k<=10) { if (l>=node[x].size()&&j>=node[y].size()) break; if (l>=node[x].size()) ans[k++]=node[y][j], ++j; else if (j>=node[y].size()) ans[k++]=node[x][l], ++l; else { ans[k++]=min(node[x][l],node[y][j]); if (node[x][l]<node[y][j]) ++l; else if (node[x][l]>node[y][j]) ++j; else ++l, ++j; } } if (dep[x]<dep[y]) swap(x,y); int t=dep[x]-dep[y]; for (int i=0; bin[i]<=t; ++i) if (bin[i]&t) merge_(x,i), x=fa[x][i][0]; for (int i=19; i>=0; --i) if (fa[x][i][0]!=fa[y][i][0]) merge_(x,i), merge_(y,i), x=fa[x][i][0], y=fa[y][i][0]; if (x==y) return ; else {merge_(x,0); return ;} } int main () { bin_init(); int n, m, q, u, v, a; scanf("%d%d%d",&n,&m,&q); FO(i,1,n) scanf("%d%d",&u,&v), add_edge(u,v), add_edge(v,u); FOR(i,1,m) scanf("%d",&u), node[u].pb(i); FOR(i,1,n) sort(node[i].begin(),node[i].end()); dfs(1,0); while (q--) { scanf("%d%d%d",&u,&v,&a); mem(ans,0); sol(u,v); int mark; for (mark=1; mark<=11&&ans[mark]; ++mark) ; mark=min(mark-1,a); printf("%d",mark); FOR(i,1,mark) printf(" %d",ans[i]); putchar(' '); } return 0; }
D.Duff in Mafia(待填坑)
E.Duff as a Queen(线段树+线性基)
给出一个数列(n<=2e5),有两种操作(q<=2e5)
1.给定区间[l,r]内的数都异或k。
2.询问区间[l,r]能够相互异或出几种数。
对于第2种操作，显然可以对区间[l,r]搞出线性基,答案就是1<<(线性基的个数).
注意到一个性质。
对于a1 a2 a3 ... an. 这些数的线性基等于 a1 a1^a2 a2^a3 ... an-1^an.的线性基。
大概就是由于这两个数列能够互相异或出来，于是能异或出来的数字种数都是相等的。
于是第一个操作就是 al^k al+1^k al+2^k ... ar^k.
对于线性基就是 al-1^a1^k a1^al+1 al+1^al+2 ... ar-1^ar ar^ar+1^k.
于是我们需要维护两个数列 一个原数列， 一个线性基与原数列相同的数列(次数列)
用BIT或者线段树维护原数列。
用线段树维护次数列的线性基。
答案即为所求。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=200005; //Code begin... int a[N], seg[N<<2][32], tree[N], n, q, ans[32]; void add(int x, int val){while (x<=n) tree[x]^=val, x+=lowbit(x);} int sum(int x){int ans=0; while (x) ans^=tree[x], x-=lowbit(x); return ans;} void push_up(int p) { mem(seg[p],0); FOR(c,0,30) seg[p][c]=seg[p<<1][c]; for (int c=30; c>=0; --c) { int x=seg[p<<1|1][c]; for (int i=c; i>=0; --i) { if (!(x>>i)) continue; if (!seg[p][i]) {seg[p][i]=x; break;} x^=seg[p][i]; } } } void bulid(int p, int l, int r) { if (l==r) { for (int c=30; c>=0; --c) if (a[l]>>c) {seg[p][c]=a[l]; break;} return ; } int mid=(l+r)>>1; bulid(lch); bulid(rch); push_up(p); } void update(int p, int l, int r, int L, int K) { if (L<l || L>r) return ; if (L==l && L==r) { a[L]^=K; mem(seg[p],0); for (int c=30; c>=0; --c) if (a[L]>>c) {seg[p][c]=a[L]; break;} } else { int mid=(l+r)>>1; update(lch,L,K); update(rch,L,K); push_up(p); } } void query(int p, int l, int r, int L, int R) { if (R<l || L>r) return ; if (L<=l && R>=r) { for (int i=30; i>=0; --i) { int x=seg[p][i]; for (int c=i; c>=0; --c) { if (!(x>>c)) continue; if (!ans[c]) {ans[c]=x; break;} x^=ans[c]; } } } else { int mid=(l+r)>>1; query(lch,L,R), query(rch,L,R); } } int main () { int flag, l, r, k; scanf("%d%d",&n,&q); FOR(i,1,n) scanf("%d",a+i); for (int i=n; i>=1; --i) a[i]^=a[i-1], add(i,a[i]); bulid(1,1,n); while (q--) { scanf("%d%d%d",&flag,&l,&r); if (flag==1) { scanf("%d",&k); update(1,1,n,l,k); add(l,k); if (r!=n) update(1,1,n,r+1,k), add(r+1,k); } else { mem(ans,0); if (l!=r) query(1,1,n,l+1,r); int temp=sum(l); for (int i=30; i>=0; --i) { if (!(temp>>i)) continue; if (!ans[i]) {ans[i]=temp; break;} temp^=ans[i]; } int cnt=0; FOR(i,0,30) if (ans[i]) ++cnt; printf("%d ",1<<cnt); } } return 0; }
F.Duff is Mad(待填坑)