题解
A AND Minimum Spanning Tree
参考代码：
#include<bits/stdc++.h> #define maxl 200010 using namespace std; int n,ans1; int mi[31]; int ans[maxl]; inline void prework() { scanf("%d",&n); } inline int find(int x) { for(int j=0;j<=30;j++) if((x&mi[j])==0) return mi[j]; } inline void mainwork() { ans1=0;int x; for(int i=2;i<=n;i++) { if(i&1) { x=find(i); if(x<=n) ans[i]=x; else ans1++,ans[i]=1; } else ans[i]=1; } } inline void print() { printf("%d ",ans1); for(int i=2;i<=n;i++) printf("%d%c",ans[i],(i==n)?' ':' '); } int main() { mi[0]=1; for(int i=1;i<=30;i++) mi[i]=mi[i-1]*2; int t; scanf("%d",&t); for(int i=1;i<=t;i++) { prework(); mainwork(); print(); } return 0; }
B Colored Tree
unsolved.
C Divide the Stones
题解：https://blog.csdn.net/liufengwei1/article/details/97970041
1 #include<bits/stdc++.h> 2 #define maxl 100010 3 using namespace std; 4 5 long long k,n,sum,t; 6 long long dy[maxl]; 7 long long last[maxl],to[maxl]; 8 long long tmp[maxl]; 9 bool flag; 10 vector <long long> ans[maxl]; 11 12 inline void prework() 13 { 14 scanf("%lld%lld",&n,&k); 15 sum=1ll*n*(n+1)/2; 16 for(long long i=1;i<=k;i++) 17 ans[i].clear(); 18 } 19 20 inline void mainwork() 21 { 22 flag=false; 23 if(sum%k!=0) 24 return; 25 sum=sum/k; 26 long long id; 27 t=n/k; 28 if(t%2==0) 29 { 30 long long id=1; 31 for(long long i=1;i<=k;i++) 32 { 33 for(long long j=1;j<=t/2;j++) 34 { 35 ans[i].push_back(id); 36 ans[i].push_back(n-id+1); 37 id++; 38 } 39 } 40 flag=true; 41 } 42 else 43 { 44 if(n/k==1) 45 { 46 if(n==1) 47 { 48 ans[1].push_back(1); 49 flag=true; 50 } 51 return; 52 } 53 for(long long i=1;i<=k/2+1;i++) 54 { 55 dy[i]=k/2+i; 56 to[i]=(i-1)*2+1; 57 } 58 for(long long i=k/2+2;i<=k;i++) 59 { 60 dy[i]=i-(k/2)-1; 61 to[i]=(i-(k/2)-1)*2; 62 } 63 for(long long i=1;i<=k;i++) 64 ans[i].push_back(i),last[i]=i,tmp[i]=i; 65 long long num; 66 for(long long i=2;i<n/k;i++) 67 { 68 for(long long j=1;j<=k;j++) 69 { 70 num=dy[last[j]]+(i-1)*k; 71 ans[j].push_back(num); 72 last[j]=to[last[j]]; 73 tmp[j]+=num; 74 } 75 } 76 for(long long i=1;i<=k;i++) 77 ans[i].push_back(sum-tmp[i]); 78 flag=true; 79 } 80 } 81 82 inline void print() 83 { 84 if(flag) 85 { 86 puts("yes"); 87 long long l; 88 for(long long i=1;i<=k;i++) 89 { 90 for(long long j=0;j<n/k;j++) 91 printf("%lld%c",ans[i][j],(j==(n/k-1))?' ':' '); 92 } 93 } 94 else 95 puts("no"); 96 } 97 98 int main() 99 { 100 long long t; 101 scanf("%lld",&t); 102 for(long long i=1;i<=t;i++) 103 { 104 prework(); 105 mainwork(); 106 print(); 107 } 108 return 0; 109 }
D Enveloping Convex
unsolved.
E Good Numbers
unsolved.
F Horse
unsolved.
G Just an Old Puzzle
题解：百度“15难题”
参考代码
#include<bits/stdc++.h> using namespace std; int ans; int a[17]; int x[17]; inline void prework() { for(int i=1;i<=16;i++) { scanf("%d",&a[i]); if(a[i]==0) { a[i]=16; ans=x[i]; } } } inline void mainwork() { for(int i=1;i<=16;i++) { for(int j=i+1;j<=16;j++) if(a[j]<a[i]) ans++; } } inline void print() { if(ans&1) puts("No"); else puts("Yes"); } int main() { x[2]=1;x[4]=1;x[5]=1;x[7]=1; x[10]=1;x[12]=1;x[13]=1;x[15]=1; int t; scanf("%d",&t); for(int cas=1;cas<=t;cas++) { prework(); mainwork(); print(); } return 0; }
H K-th Closest Distance
题解：主席树+二分 https://blog.csdn.net/liufengwei1/article/details/97948584
参考代码
#include<bits/stdc++.h> #define maxl 100010 using namespace std; const int nn=1e6; int n,m,tot,ans; int rt[maxl],a[maxl]; struct node { int ls,rs,sum; }tree[maxl*10*35]; inline void insert(int num,int &x,int l,int r) { tree[++tot]=tree[x];x=tot; ++tree[x].sum; if(l==r) return; int mid=(l+r)>>1; if(num<=mid) insert(num,tree[x].ls,l,mid); else insert(num,tree[x].rs,mid+1,r); } inline void prework() { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&a[i]); tree[0].ls=tree[0].rs=tree[0].sum=0; rt[0]=0; tot=0; for(int i=1;i<=n;i++) { rt[i]=rt[i-1]; insert(a[i],rt[i],1,nn); } } inline int query(int i,int j,int l,int r,int i1,int j1) { if(i1==l && j1==r) return tree[j].sum-tree[i].sum; int mid=(i1+j1)>>1,ret; if(r<=mid) ret=query(tree[i].ls,tree[j].ls,l,r,i1,mid); else if(l>mid) ret=query(tree[i].rs,tree[j].rs,l,r,mid+1,j1); else { ret=query(tree[i].ls,tree[j].ls,l,mid,i1,mid); ret+=query(tree[i].rs,tree[j].rs,mid+1,r,mid+1,j1); } return ret; } inline bool jug(int l,int r,int mid,int p,int k) { int up=min(p+mid,nn),lo=max(1,p-mid); int sum=query(rt[l-1],rt[r],lo,up,1,nn); if(sum<k) return false; else return true; } inline void mainwork() { ans=0;int up,lo,p,k,l,r,mid; for(int i=1;i<=m;i++) { scanf("%d%d%d%d",&lo,&up,&p,&k); lo^=ans;up^=ans;p^=ans;k^=ans; l=0;r=nn; while(l+1<r) { mid=(l+r)>>1; if(!jug(lo,up,mid,p,k))// <k l=mid; else r=mid; } if(jug(lo,up,l,p,k)) ans=l; else ans=l+1; printf("%d ",ans); } } inline void print(){} int main() { int t; scanf("%d",&t); for(int i=1;i<=t;i++) { prework(); mainwork(); print(); } return 0; }
I Linear Functions
unsolved.
J Minimal Power of Prime
参考代码
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int size=1e6+5; double eps=1e-8; int p[size];bool prime[size]; int mpri[size]; int tot=0; void init() { for(int i=1;i<size;i++) prime[i]=true; for(int i=2;i<size;i++) { if(prime[i]) { p[++tot]=i; mpri[i]=i; } for(int j=1;j<=tot&&p[j]*i<size;j++) { prime[i*p[j]]=false;mpri[i*p[j]]=p[j]; if(i%p[j]==0) break; } } } int main() { init(); int t; long long x; scanf("%d",&t); while(t--) { scanf("%lld",&x); int cnt=0; int ans=64; if(x<size){ int ps=mpri[x]; while(x!=1) { do x/=ps,cnt++; while(mpri[x]==ps); ps=mpri[x]; ans=min(ans,cnt); } printf("%d ",ans); continue; } bool flag=false; for(int i=1;i<=tot;i++) { cnt=0; if(x%p[i]==0) { do x/=p[i],cnt++; while(x%p[i]==0); } if(cnt==1) { puts("1"); flag=true; break; } } if(flag) continue; LL sq=sqrt(x)+eps; if(sq*sq==x) { printf("2 "); } else { puts("1"); } } }