9.25模拟赛
作者：互联网
1 #include<bits/stdc++.h>
2 #define ll long long
3 using namespace std;
4 const int N=2e6+11,M=61,L=(1<<13)+11,St=(1<<13);
5 int n,a[M];
6 int tot,fr[N],to[N],g[L][L];
7 ll dp[2][N];
8
9 inline int re_ad() {
10 char ch=getchar(); int x=0,f=1;
11 while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
12 while('0'<=ch && ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
13 return x*f;
14 }
15
16 int main()
17 {
18 n=re_ad();
19 for(int i=1;i<=n;++i) a[i]=re_ad();
20 for(int i=0;i<St;++i) {
21 int ii=i^(St-1);
22 for(int j=ii;j;j=(j-1)&ii) g[j][i]=++tot,fr[tot]=i,to[tot]=j;
23 g[0][i]=++tot,fr[tot]=i,to[tot]=0;
24 }
25 // printf("%d\n",tot);
26 // for(int i=1;i<=20;++i) printf("%d %d\n",fr[i],to[i]);
27 dp[0][g[0][0]]=1;
28 // printf("g[0][0]: %d\n",g[0][0]);
29 for(int i=1;i<=n;++i) {
30 int t=(i&1)^1;
31 for(int j=1;j<=tot;++j) dp[t^1][j]=dp[t][j];
32 for(int j=1;j<=tot;++j) {
33 if(!dp[t][j]) continue;
34 // printf("%d %d: %d %d\n",i,j,fr[j]|to[j],to[j]);
35 int x=fr[j]|to[j],y=fr[j];
36 x|=a[i],y^=a[i];
37 dp[t^1][g[x^y][y]]+=dp[t][j];
38 }
39 }
40 ll ans=0;
41 for(int i=1;i<=tot;++i) if(!to[i] && fr[i]) ans+=dp[n&1][i];
42 printf("%lld\n",ans);
43 return 0;
44 }
1 #include<bits/stdc++.h>
2 using namespace std;
3 const int N=3e5+11,M=61,St=50;
4 int n,m,a[N];
5
6 inline int re_ad() {
7 char ch=getchar(); int x=0,f=1;
8 while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
9 while('0'<=ch && ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
10 return x*f;
11 }
12
13 struct Node {
14 int len,val[M];
15 Node Init() { len=0; for(int i=1;i<=St;++i) val[i]=0; }
16 Node operator + (const Node &B) const {
17 int d1=1,d2=1;
18 Node res; res.len=len+B.len;
19 for(int i=1;i<=St;++i) {
20 if(d1>len && d2>B.len) return res;
21 if(d1>len) res.val[i]=B.val[d2++];
22 else if(d2>B.len) res.val[i]=val[d1++];
23 else if(val[d1]>B.val[d2]) res.val[i]=val[d1++];
24 else res.val[i]=B.val[d2++];
25 }
26 return res;
27 }
28 };
29
30 namespace ST {
31 Node t[N];
32 inline void Pushup(int d) {
33 t[d]=t[d<<1]+t[d<<1|1];
34 }
35 void build(int d,int l,int r) {
36 t[d].Init(); t[d].len=r-l+1;
37 if(l==r) { t[d].val[1]=a[l]; return ; }
38 int mid=(l+r)>>1;
39 build(d<<1,l,mid);
40 build(d<<1|1,mid+1,r);
41 Pushup(d);
42 }
43 inline void Build() { build(1,1,n); }
44 void modify(int d,int l,int r,int x,int z) {
45 if(l==r) { t[d].val[1]=z; return ; }
46 int mid=(l+r)>>1;
47 (x<=mid)?modify(d<<1,l,mid,x,z):modify(d<<1|1,mid+1,r,x,z);
48 Pushup(d);
49 }
50 inline void Modify(int x,int z) { modify(1,1,n,x,z); }
51 Node query(int d,int l,int r,int x,int y) {
52 Node res; res.Init();
53 if(x>y || l>r) return res;
54 if(x<=l && r<=y) return t[d];
55 int mid=(l+r)>>1;
56 if(x<=mid) res=res+query(d<<1,l,mid,x,y);
57 if(y>=mid+1) res=res+query(d<<1|1,mid+1,r,x,y);
58 return res;
59 }
60 inline void Query(int x,int y) {
61 Node tmp=query(1,1,n,x,y);
62 // printf("query: %d %d: %d: %d %d %d %d %d\n",x,y,tmp.len,tmp.val[1],tmp.val[2],tmp.val[3],tmp.val[4],tmp.val[5]);
63 if(tmp.len<3) { printf("0\n"); return ; }
64 for(int i=1;i<=tmp.len-2;++i) if(tmp.val[i]<tmp.val[i+1]+tmp.val[i+2])
65 { printf("%d\n",tmp.val[i]+tmp.val[i+1]+tmp.val[i+2]); return ; }
66 printf("0\n");
67 }
68 /*
69 void print_search(int d,int l,int r) {
70 printf("%d %d %d: %d: %d %d %d %d %d\n",d,l,r,t[d].len,t[d].val[1],t[d].val[2],t[d].val[3],t[d].val[4],t[d].val[5]);
71 if(l==r) return ;
72 int mid=(l+r)>>1;
73 print_search(d<<1,l,mid),print_search(d<<1|1,mid+1,r);
74 }
75 inline void Print_Test() {
76 print_search(1,1,n);
77 }
78 */
79 }
80
81 int main()
82 {
83 n=re_ad(),m=re_ad();
84 for(int i=1;i<=n;++i) a[i]=re_ad();
85 ST::Build();
86 // ST::Print_Test();
87 for(int i=1,op,x,y;i<=m;++i) {
88 op=re_ad(),x=re_ad(),y=re_ad();
89 // if(op==2) ST::Query(x,y);
90 op==1 ? ST::Modify(x,y) : ST::Query(x,y);
91 // ST::Print_Test();
92 }
93 return 0;
94 }
1 //std
2 #include <cstdio>
3 #include <algorithm>
4 #include<iostream>
5 #include <cstring>
6 using namespace std;
7 const int MAX_N = 1000005, S = (1 << 20) - 1;
8 class Edge
9 {
10 public:
11 int nxt, to;
12 } e[MAX_N << 1];
13 int head[MAX_N], cnt;
14 void addedge(int u, int v)
15 {
16 e[++cnt] = (Edge){head[u], v}, head[u] = cnt;
17 e[++cnt] = (Edge){head[v], u}, head[v] = cnt;
18 }
19 int N, dep[MAX_N], f[1 << 20];
20 void dfs(int u, int v)
21 {
22 dep[u] = dep[v] + 1;
23 for (int i = head[u]; i; i = e[i].nxt)
24 if (e[i].to != v)
25 dfs(e[i].to, u);
26 }
27 int val[MAX_N];
28 int main()
29 {
30 freopen("tree.in","r",stdin);
31 freopen("tree.out","w",stdout);
32 scanf("%d", &N);
33
34 for (int i = 1; i <= N; ++i)
35 scanf("%d", &val[i]);
36 for (int i = 1, u, v; i < N; ++i)
37 {
38 scanf("%d%d", &u, &v);
39 addedge(u, v);
40 }
41 dfs(1, 0);
42
43 for (int i = 1; i <= N; ++i) f[dep[i] - 1] ^= val[i];
44
45
46
47 for (int i = 0; i < 20; ++i)
48 for (int j = 0; j <= S; ++j)
49 if ((j >> i) & 1)
50 f[j] ^= f[j ^ (1 << i)];
51
52
53
54 for (int i = 0; i < N; ++i)
55 printf("%d ", f[i ^ S]);
56 puts("");
57 return 0;
58 }
标签：ch,9.25,val,int,res,include,d2,模拟 来源： https://www.cnblogs.com/shangguanshufang/p/9_25_mo_ni_sai.html