2017-10-04-afternoon

注意完全平方数统计时的特判

 1 #include <cstdio>
 2 
 3 inline void read(int &x)
 4 {
 5     x=0; register char ch=getchar();
 6     for(; ch>'9'||ch<'0'; ) ch=getchar();
 7     for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
 8 }
 9 const int N(40005);
10 int n,cnt[N],ans;
11 
12 inline void Update(int x)
13 {
14     int i;
15     for(i=1; i*i<x; ++i)
16         if(x%i==0) cnt[i]++,cnt[x/i]++;
17     if(i*i==x) cnt[i]++;
18 }
19 
20 int Presist()
21 {
22     freopen("a.in","r",stdin);
23     freopen("a.out","w",stdout);
24 
25     read(n);
26     for(int op,x; n--; )
27     {
28         read(op),read(x);
29         if(op==1) Update(x);
30         else ans^=cnt[x];
31     }
32     printf("%d
",ans);
33     return 0;
34 }
35 
36 int Aptal=Presist();
37 int main(int argc,char**argv){;}

 1 #include <cstring>
 2 #include <cstdio>
 3 
 4 #define LL long long
 5 inline void read(int &x)
 6 {
 7     x=0; register char ch=getchar();
 8     for(; ch>'9'||ch<'0'; ) ch=getchar();
 9     for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
10 }
11 const int N(3e5+5);
12 
13 int cnt[2][N],val[N],dad[N];
14 int sumedge,head[N];
15 struct Edge {
16     int v,next,w;
17     Edge(int v=0,int next=0,int w=0):v(v),next(next),w(w){}
18 }edge[N<<1];
19 inline void ins(int u,int v,int w)
20 {
21     cnt[0][u]++; cnt[0][v]++;
22     edge[++sumedge]=Edge(v,head[u],w); head[u]=sumedge;
23     edge[++sumedge]=Edge(u,head[v],w); head[v]=sumedge;
24 }
25 
26 bool vis[N];
27 long long ans;
28 
29 inline void Add(int u)
30 {
31     for(; dad[u]; u=dad[u]) cnt[1][dad[u]]++;
32 }
33 void DFS(int u,int fa,int pre)
34 {
35     for(int v,i=head[u]; i; i=edge[i].next)
36     {
37         v=edge[i].v;
38         if(vis[v]||fa==v||edge[i].w==pre) continue;
39         cnt[1][v]++, dad[v]=u;
40         DFS(v,u,edge[i].w); Add(v); dad[v]=0;
41     }
42 }
43 
44 int Presist()
45 {
46     freopen("b.in","r",stdin);
47     freopen("b.out","w",stdout);
48     int n; read(n);
49     for(int i=1; i<=n; ++i) read(val[i]);
50     for(int u,v,w,i=1; i<n; ++i)
51         read(u),read(v),read(w),ins(u,v,w);
52     for(int i=1; i<=n; ++i)
53         if(cnt[0][i]==1) vis[i]=1,DFS(i,0,0);
54     for(int i=1; i<=n; ++i)
55         if(!vis[i]) vis[i]=1,DFS(i,0,0);
56     for(int i=1; i<=n; ++i)
57         ans+=1ll*cnt[1][i]*val[i];
58     printf("%I64d
",ans);
59     return 0;
60 }
61 
62 int Aptal=Presist();
63 int main(int argc,char**argv){;}

额。。想法有问题

1 /*
2 呃呃呃
3 */

未完成

——每当你想要放弃的时候，就想想是为了什么才一路坚持到现在。