The exam again expose my big problem.
First, it is better to do a few times a minute examination did not hang,
But it also reflects the larger problem, and that is my strength seems to be limited to the.
Examination acquire full of violence points (100 + 0 + 50), and then looked at T2 did not understand, I called memory search, only 10 points, this time only after less than two hours
Did not understand the question it when it really is my food, but then, I did not go to think deeply cut on the T3, skyh exam questions, I do not have even the most basic ideas,
This problem may be on the first question, I would cut.
This is not a problem, I know it is the exam after a inclusion-exclusion, also launched its own here.
But take only part of the sub-T3 thinking imprisoned me, the results of my own thinking aimlessly, but always in vain.
In the latter part of the exam just because I do not have the pressure, thinking Wenna of 150 points is enough.
Really. . . enough?
Effective think time test positive correlation with the results.
I can allow myself to burst zero, but I do not like doing nothing the whole exam.
Stronger than you, normal examination are thinking, and you're ahead of the Resurrection, which is the reason why you are better than the reasons for it.
Maybe you have a fairly good-looking ranking, so what?
To be honest, I remember the construction of the city that question, not counting water, brush if on the usual question, a half hour I could not do it, but in the examination room, I did cut off.
This is the power of so-called under the pressure of it.
A person can accept failure, but it must not stoop.
You are not strong enough, this is true.
But if you have fear of difficulty, you can never strong.
You're not a genius, but they can not wallow in mediocrity.
You still need to experience
Said exam
After T1 DP
T2 is intended to convert title -> Given a grid, all paths come to ask boundaries maximum value from each of the smallest grid.
Minimum spanning tree to run.
1 #include<bits/stdc++.h> 2 #define int long long 3 using namespace std; 4 const int root=1000005; 5 const int tmpx[5]={0,0,1,-1},tmpy[5]={1,-1,0,0}; 6 struct edge{ 7 int st,ed,val; 8 bool ok; 9 friend bool operator < (const edge a,const edge b) 10 { 11 return a.val<b.val; 12 } 13 }E[1000040]; 14 int cnt,head[1000040],nxt[1000040],to[1000040],pnt,f[1000040],w[1000040],n,m; 15 int d[1000040],h[304][305]; 16 int pt(int x,int y) 17 { 18 if(!x||!y||x==n+1||y==m+1)return root; 19 return (x-1)*m+y; 20 } 21 void Add(int u,int v,int val) 22 { 23 to[++pnt]=v; 24 nxt[pnt]=head[u]; 25 w[pnt]=val; 26 head[u]=pnt; 27 return ; 28 } 29 int find(int x) 30 { 31 if(f[x]==x)return x; 32 else return f[x]=find(f[x]); 33 } 34 void merge(int a,int b) 35 { 36 int fa=find(a),fb=find(b); 37 f[fa]=fb; 38 return ; 39 } 40 void dfs(int x,int fa) 41 { 42 for(int i=head[x];i;i=nxt[i]) 43 { 44 int y=to[i]; 45 if(y==fa)continue; 46 d[y]=max(d[x],w[i]); 47 dfs(y,x); 48 } 49 return ; 50 } 51 void BuildAndWork() 52 { 53 for(int i=1;i<=root;i++)f[i]=i; 54 for(int i=1;i<=n;i++) 55 for(int j=1;j<=m;j++) 56 { 57 for(int k=0;k<4;k++) 58 { 59 E[++cnt].st=pt(i,j); 60 E[cnt].ed=pt(i+tmpx[k],j+tmpy[k]); 61 E[cnt].val=max(h[i][j],h[i+tmpx[k]][j+tmpy[k]]); 62 } 63 } 64 sort(E+1,E+cnt+1); 65 for(int i=1;i<=cnt;i++) 66 { 67 int a=E[i].st,b=E[i].ed; 68 if(find(a)==find(b))continue; 69 Add(a,b,E[i].val);Add(b,a,E[i].val); 70 merge(a,b); 71 } 72 dfs(root,0); 73 } 74 signed main() 75 { 76 scanf("%lld%lld",&n,&m); 77 for(int i=1;i<=n;i++) 78 for(int j=1;j<=m;j++) 79 scanf("%lld",&h[i][j]); 80 BuildAndWork(); 81 int p=0; 82 for(int i=1;i<=n;i++) 83 { 84 for(int j=1;j<=m;j++) 85 { 86 int now=d[++p]; 87 printf("%lld ",now-h[i][j]); 88 } 89 puts(""); 90 } 91 return 0; 92 }
T3 I want to say.
On the test data structure always thought it was a problem.
But this is relatively prime ah, why not think about the mathematics of it?
This is just a simple inclusion-exclusion ah.
Think of inclusion and exclusion, everything will be solved.
Think of it? Ha ha.
Direct inclusion and exclusion can be.
O (msqrt (MAX {x [i]})) can be updated.
1 #include<bits/stdc++.h> 2 #define MAXN 500005 3 #define cri const rigister int 4 using namespace std; 5 int t[MAXN],firprime[MAXN]; 6 int prime[MAXN],x[MAXN],val[MAXN],bin[MAXN],bitnum[MAXN]; 7 bool vst[MAXN]; 8 int lm[10]; 9 int kx; 10 void pre() 11 { 12 for(register int i=2;i<=500000;i++) 13 { 14 if(!firprime[i]) 15 prime[++prime[0]]=i,firprime[i]=i; 16 for(register int j=1;j<=prime[0]&&i*prime[j]<=500000;j++) 17 { 18 firprime[i*prime[j]]=prime[j]; 19 if(!(i%prime[j]))break; 20 } 21 } 22 return ; 23 } 24 void Get(register int x) 25 { 26 lm[0]=0; 27 while(x!=1) 28 { 29 register int now=firprime[x]; 30 lm[++lm[0]]=now; 31 while(x%now==0)x/=now; 32 } 33 return ; 34 } 35 int main() 36 { 37 pre(); 38 int n,m,siz=0; 39 long long ans=0; 40 val[0]=1; 41 for(int i=1;i<=7;i++)bin[1<<(i-1)]=i; 42 for(int i=1;i<=(1<<7-1);i++) 43 { 44 int num=0,st=i; 45 while(st)st-=st&-st,++num; 46 bitnum[i]=num; 47 } 48 scanf("%d%d",&n,&m); 49 for(int i=1;i<=n;i++) 50 scanf("%d",&x[i]); 51 while(m--) 52 { 53 int opt; 54 scanf("%d",&opt); 55 if(!vst[opt]) 56 { 57 Get(x[opt]); 58 kx=siz; 59 for(register int i=1;i<=(1<<lm[0])-1;i++) 60 { 61 val[i]=val[i^(i&-i)]*lm[bin[i&-i]]; 62 if(bitnum[i]&1)kx-=t[val[i]]; 63 else kx+=t[val[i]]; 64 } 65 vst[opt]=1; 66 int sx=sqrt(x[opt]); 67 for(register int i=2;i<=sx;i++) 68 if(x[opt]%i==0) 69 { 70 if(i*i==x[opt])t[i]++; 71 else t[i]++,t[x[opt]/i]++; 72 } 73 t[x[opt]]++; 74 siz++; 75 ans+=kx; 76 } 77 else 78 { 79 Get(x[opt]); 80 vst[opt]=0; 81 int sx=sqrt(x[opt]); 82 for(register int i=2;i<=sx;i++) 83 if(x[opt]%i==0) 84 { 85 if(i*i==x[opt])t[i]--; 86 else t[i]--,t[x[opt]/i]--; 87 } 88 t[x[opt]]--;siz--; 89 kx=siz; 90 for(register int i=1;i<=(1<<lm[0])-1;i++) 91 { 92 val[i]=val[i^(i&-i)]*lm[bin[i&-i]]; 93 if(bitnum[i]&1)kx-=t[val[i]]; 94 else kx+=t[val[i]]; 95 } 96 ans-=kx; 97 } 98 printf("%lld\n",ans); 99 } 100 return 0; 101 }
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