Day2 还是很有难度的emmmm
A - Por Costel and Azerah
给定n个数，从其中取出若干个数使其和为偶数，问有多少种取法
（1）dp做法：挖坑待更
（2）数学做法：可以分为三种情况：
1，从奇数中取2*x个 s1=C(n,2)+C(n,4)+...+C(n,n/2*2)
2，从偶数中取任意个s2=C(n,1)+C(n,2)+C(n,3)+...+C(n,n)
3，从奇数中取2*x个，偶数中取任意个s3=s1*s2
用组合数可以算出来
1 #include<bits/stdc++.h> 2 using namespace std; 3 int odd,even; 4 const int mod=1e9+7; 5 int power(int a,int n) 6 { 7 int ans=1,tmp=a; 8 while(n) 9 { 10 //cout<<ans<<endl; 11 if(n&1) ans=1ll*ans*tmp%mod; 12 tmp=1ll*tmp*tmp%mod; 13 n/=2; 14 } 15 return ans; 16 } 17 int main() 18 { 19 freopen("azerah.in","r",stdin); 20 freopen("azerah.out","w",stdout); 21 ios::sync_with_stdio(0); 22 int n,t;cin>>t; 23 while(t--){ 24 even=odd=0; 25 cin>>n; 26 for(int i=1;i<=n;++i) 27 { 28 int t;cin>>t; 29 if(t&1) ++odd; 30 else ++even; 31 } 32 int s1=power(2,odd-1)-1,s2=power(2,even)-1; 33 if(odd==0) s1=0;if(even==0) s2=0; 34 long long ans=(s1+s2+1ll*s1*s2)%mod; 35 cout<<ans<<endl; 36 } 37 }
B - Por Costel and the Algorithm
参考https://www.cnblogs.com/autsky-jadek/p/6338115.html
给一个有向图，求它的最短路树，先输出最短路树的边，再输出其他边
1 #include<bits/stdc++.h> 2 #define ll long long 3 #define pil pair<int,ll> 4 #define mem(a) memset(a,0,sizeof(a)) 5 using namespace std; 6 const int maxn=2e5+5; 7 ll dis[maxn]; 8 int fir[maxn],pre[maxn],n,m,cnt; 9 bool vis[maxn],used[maxn*2]; 10 struct edge { int to,v,nex;}E[maxn*2]; 11 struct cmp { bool operator()(pil a,pil b){return a.second>b.second;}}; 12 void add(int fro,int to,int v){E[++cnt].to=to,E[cnt].v=v,E[cnt].nex=fir[fro],fir[fro]=cnt;} 13 void ini(){mem(fir),mem(pre),mem(vis),mem(E),mem(used);fill(dis,dis+maxn,1e18);cnt=0;} 14 void dfs(int now) 15 { 16 for(int i=fir[now];i;i=E[i].nex) 17 { 18 if(used[i]) 19 { 20 cout<<i<<" "; 21 dfs(E[i].to); 22 } 23 } 24 } 25 void dij() 26 { 27 priority_queue<pil,vector<pil>,cmp>pq; 28 pq.push({1,0}); 29 dis[1]=0; 30 while(!pq.empty()) 31 { 32 int now=pq.top().first; 33 pq.pop(); 34 if(vis[now]) continue; 35 vis[now]=1; 36 for(int i=fir[now];i;i=E[i].nex) 37 { 38 int to=E[i].to, 39 v=E[i].v; 40 if(vis[to]==0 && dis[to]>dis[now]+v) 41 { 42 //cout<<"test:"<<endl; 43 //printf("i: %d now: %d to:%d used: %d pre: %d\n",i,now,to,used[pre[to]],pre[to]); 44 dis[to]=dis[now]+v; 45 used[pre[to]]=0; 46 used[i]=1; 47 pre[to]=i; 48 pq.push({to,dis[to]}); 49 } 50 } 51 } 52 } 53 int main() 54 { 55 freopen("algoritm.in","r",stdin); 56 freopen("algoritm.out","w",stdout); 57 ios::sync_with_stdio(0); 58 int t;cin>>t; 59 while(t--) 60 { 61 ini(); 62 cin>>n>>m; 63 for(int i=1;i<=m;++i) 64 { 65 int a,b,c; cin>>a>>b>>c; 66 add(a,b,c); 67 } 68 dij(); 69 dfs(1); 70 for(int i=1;i<=m;++i) if(!used[i]) cout<<i<<" "; 71 cout<<endl; 72 } 73 }
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define mem(a) memset(a,0,sizeof(a)) 4 using namespace std; 5 const int maxn=1e5+5; 6 struct pii{int to,v,id;}; 7 struct pil{int first;ll second;}; 8 struct cmp{bool operator()(pil a,pil b){return a.second>b.second;}}; 9 vector<pii>G[maxn]; 10 int pre[maxn],n,m; 11 ll dis[maxn]; 12 bool vis[maxn],used[maxn*2]; 13 void ini() { mem(pre),mem(vis),mem(used);for(int i=0;i<maxn;++i)G[i].clear();fill(dis,dis+maxn,1e18);} 14 void dij() 15 { 16 priority_queue<pil,vector<pil>,cmp>pq; 17 pq.push({1,0}); 18 dis[1]=0; 19 while(!pq.empty()) 20 { 21 int now=pq.top().first; 22 pq.pop(); 23 if(vis[now]) continue;vis[now]=1; 24 for(auto t:G[now]) 25 { 26 int to=t.to,v=t.v,id=t.id; 27 if(!vis[to] && dis[to]>dis[now]+v) 28 { 29 dis[to]=dis[now]+v; 30 pq.push({to,dis[to]}); 31 used[pre[to]]=0; 32 pre[to]=id; 33 used[id]=1; 34 } 35 } 36 } 37 } 38 void dfs(int now) 39 { 40 for(auto t:G[now]) 41 { 42 int to=t.to; 43 if(used[t.id]) 44 { 45 cout<<t.id<<" "; 46 dfs(to); 47 } 48 } 49 } 50 int main() 51 { 52 freopen("algoritm.in","r",stdin); 53 freopen("algoritm.out","w",stdout); 54 ios::sync_with_stdio(0); 55 int t;cin>>t; 56 while(t--) { 57 ini(); 58 cin >> n >> m; 59 for (int i = 1; i <= m; ++i) { 60 int a, b, c; 61 cin >> a >> b >> c; 62 G[a].push_back({b, c, i}); 63 } 64 dij(); 65 dfs(1); 66 for(int i=1;i<=m;++i) if(used[i]==0) cout<<i<<" ";cout<<endl; 67 } 68 }
H - Por Costel and the Match
分类并查集，参考poj食物链
还有一个做法是带权并查集不过先占个坑
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=2e5+5; 4 int fa[maxn]; 5 int fin(int x){return x==fa[x]?x:fa[x]=fin(fa[x]);} 6 7 int main() 8 { 9 freopen("meciul.in","r",stdin); 10 freopen("meciul.out","w",stdout); 11 int t;scanf("%d",&t); 12 while(t--) 13 { 14 int n,m; 15 scanf("%d%d",&n,&m); 16 for(int i=1;i<maxn;++i) fa[i]=i; 17 for(int i=1;i<=m;++i) 18 { 19 int a,b; 20 scanf("%d%d",&a,&b); 21 int x=fin(a),y=fin(b),c=fin(a+n); 22 if(x==y) puts("NO"); 23 else{ 24 puts("YES"); 25 a=fin(a+n),b=fin(b+n); 26 fa[y]=a; 27 fa[x]=b; 28 } 29 } 30 } 31 }
I - Por Costel and the Pairs
数形结合一下？画个y=k/x的图像，用面积搞一下
（牛逼网友说可以分块打表emmm）
1 #include<bits/stdc++.h> 2 #define int long long 3 using namespace std; 4 5 signed main() 6 { 7 freopen("perechi3.in","r",stdin); 8 freopen("perechi3.out","w",stdout); 9 ios::sync_with_stdio(0); 10 int t;cin>>t; 11 while (t--) 12 { 13 int n;cin>>n; 14 int i; 15 int ans=0; 16 for(i=1;i*i<=n;++i) 17 { 18 ans+=n/i; 19 } 20 ans=ans*2-(--i)*i; 21 cout<<ans<<endl; 22 } 23 }
K - Por Costel and the Firecracker
根据mod，可知要求的数下标在1e7级别，不过限制了内存，所以可以分块打表，先隔1000个打表一次，可以得到一个1e4的表，对于每次询问，可以在1000次之内得到结果
（也可以隔100个打表一次，合理即可）注意下标要从0开始
这是隔100的
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e5+5; 4 const int mod=1e7+3; 5 int pos[maxn]; 6 int main() 7 { 8 freopen("pocnitoare.in","r",stdin); 9 freopen("pocnitoare.out","w",stdout); 10 ios::sync_with_stdio(0); 11 int n,a,b,x,q,q1; 12 13 int ans=0; 14 cin>>n>>a>>b>>x>>q>>q1; 15 int xx=x; 16 pos[0]=x-a; 17 for(int i=1;i<=10000000;++i) 18 { 19 if(i%100==0) 20 { 21 pos[i/100]=x; 22 } 23 if(i==q1) ans=x; 24 x=(1ll*i*x+a)%n; 25 } 26 cout<<ans<<endl; 27 int tmp=ans; 28 for(int i=2;i<=q;++i) 29 { 30 int qnow=((1ll*tmp*(i-1)+b)+1)%mod; 31 int start=qnow/100; 32 tmp=pos[start]; 33 for(int j=start*100;j<qnow;++j) 34 { 35 tmp=(1ll*j*tmp+a)%n; 36 if(j==0) tmp=xx; 37 } 38 cout<<tmp<<endl; 39 } 40 }
L - Por Costel and the Semipalindromes
定义半回文串是一定长度的前缀和后缀相等的字符串，给定n和k，求一个长度为n的字典序第k小的半回文串
前缀/后缀长度取1即可
把k分解为长度为n-1的01串即可，并使s[n]=s[1]
1 #include<bits/stdc++.h> 2 #define int long long 3 using namespace std; 4 int n,k; 5 vector<int> div(int nn) 6 { 7 vector<int>v; 8 while(nn) 9 { 10 v.push_back(nn&1); 11 nn/=2; 12 } 13 while (v.size()<n-1) 14 { 15 v.push_back(0); 16 } 17 reverse(v.begin(),v.end()); 18 return v; 19 } 20 signed main() 21 { 22 freopen("semipal.in","r",stdin); 23 freopen("semipal.out","w",stdout); 24 ios::sync_with_stdio(0); 25 int t;cin>>t; 26 while(t--) 27 { 28 cin>>n>>k; 29 vector<int>v; 30 v=div(k-1); 31 v.push_back(v[0]); 32 for(int i:v)cout<<(i?'b':'a'); 33 cout<<endl; 34 } 35 }
睡了，待更。。。（咕了，以后没了）