A. Filling Diamonds
F(n)=F(n-1)+1
B. Sorted Adjacent Differences
最左边，然后最右边，往返取
C. Powered Addition
假设二分的数量的K，从前往后，若当前数比前面的最大值小，那么有两种情况：
- 2^K-1也就是说有的都填了，还不够，此时返回false
- 可以，那么一定可以刚好布满
/*
* Author : Jk_Chen
* Date : 2020-04-17-10.55.11
*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
void test(){cerr<<"\n";}
template<typename T,typename... Args>void test(T x,Args... args){cerr<<x<<" ";test(args...);}
const LL mod=1e9+7;
const int maxn=1e5+9;
const int inf=0x3f3f3f3f;
LL rd(){ LL ans=0; char last=' ',ch=getchar();
while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
#define rd rd()
/*_________________________________________________________begin*/
int n;
LL a[maxn];
bool check(int mx){
LL pre=a[1];
rep(i,2,n){
if(a[i]>=pre){pre=a[i];continue;}
LL need=pre-a[i];
if((1ll<<mx)-1<need)return 0;
pre=max(pre,a[i]);
}
return 1;
}
int main(){
int t=rd;
while(t--){
n=rd;
rep(i,1,n)a[i]=rd;
int l=0,r=35,ans=-1;
while(l<=r){
int mid=l+r>>1;
if(check(mid))ans=mid,r=mid-1;
else l=mid+1;
}
printf("%d\n",ans);
}
return 0;
}
/*_________________________________________________________end*/
D. Edge Weight Assignment
选一个非叶子作为root。
一个结点的儿子有多个叶子时，最大值减去叶子数-1
最小值，当存在两个儿子一奇一偶则为3，否则为1
/*
* Author : Jk_Chen
* Date : 2020-04-17-11.09.01
*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
void test(){cerr<<"\n";}
template<typename T,typename... Args>void test(T x,Args... args){cerr<<x<<" ";test(args...);}
const LL mod=1e9+7;
const int maxn=1e5+9;
const int inf=0x3f3f3f3f;
LL rd(){ LL ans=0; char last=' ',ch=getchar();
while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
#define rd rd()
/*_________________________________________________________begin*/
#define rep_e(i,p,u) for(int i=head[p],u=to[i];i;i=nex[i],u=to[i])
int head[maxn],to[maxn<<1],nex[maxn<<1],now;
void add(int a,int b){
nex[++now]=head[a];head[a]=now;to[now]=b;
}
void init_edge(){
memset(head,0,sizeof head);
now=0;
}
/*_________________________________________________________edge*/
int siz[maxn],flag[maxn];
int ans1,ans2;
void dfs(int p,int fa){
siz[p]=1;
flag[p]=1;
int sonstate=0;
int numleave=0;
rep_e(i,p,u){
if(u==fa)continue;
dfs(u,p);
siz[p]+=siz[u];
if(siz[u]==1)numleave++;
sonstate|=flag[u];
}
if(numleave>1)ans2-=numleave-1;
if(sonstate==3)ans1=3;
else if(sonstate==1)flag[p]=2;
}
int deg[maxn];
int main(){
int n=rd;
rep(i,1,n-1){
int a=rd,b=rd;
add(a,b);add(b,a);
deg[a]++;deg[b]++;
}
ans1=1,ans2=n-1;
int rt;
rep(i,1,n)if(deg[i]>1){rt=i;break;}
dfs(rt,0);
printf("%d %d\n",ans1,ans2);
return 0;
}
/*_________________________________________________________end*/
E. Perfect Triples
打表，发现对于每3个数，第一个数是1,4-7,16-31,-64-127…
分析每一组的第一个数在二进制上，每2位为一个单位，00|00|00变为01|10|11变为10|11|01再变为11|01||10
我们得到这一组的第一个数，然后推出第2个和第3个
/*
* Author : Jk_Chen
* Date : 2020-04-17-11.30.35
*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
void test(){cerr<<"\n";}
template<typename T,typename... Args>void test(T x,Args... args){cerr<<x<<" ";test(args...);}
const LL mod=1e9+7;
const int maxn=1e5+9;
const int inf=0x3f3f3f3f;
LL rd(){ LL ans=0; char last=' ',ch=getchar();
while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
#define rd rd()
/*_________________________________________________________begin*/
namespace check{
vector<int>ans;
bool vis[maxn];
void find(int p){
vis[p]=1;
rep(i,p+1,10000){
if(vis[i])continue;
if(vis[p^i])continue;
vis[i]=1;vis[p^i]=1;
ans.pb(p);
ans.pb(i);
ans.pb(p^i);
return;
}
}
void Main(){
rep(i,1,1000){
if(!vis[i])find(i);
}
int i=0;
for(auto P:ans){
i++;
bitset<15>num(P);
cout<<num<<" "<<P<<endl;
if(i%3==0)printf("_____%d________\n",i/3);
}
}
}
void find(LL n){
LL zu=(n+2)/3;
LL mul=1;
while(zu>mul){
zu-=mul;
mul*=4;
}
LL fi=mul-1+zu;
// printf("fi=%lld\n",fi);
bitset<63>Z(fi);
LL index=(n-1)%3;
rep(i,0,61){
if(Z[i]==1&&Z[i+1]==0){
if(index==1){
Z.set(i+1);
Z.reset(i);
}
else if(index==2){
Z.set(i);
Z.set(i+1);
}
}
else if(Z[i]==0&&Z[i+1]==1){
if(index==1){
Z.set(i);
Z.set(i+1);
}
else if(index==2){
Z.set(i);
Z.reset(i+1);
}
}
else if(Z[i]==1&&Z[i+1]==1){
if(index==1){
Z.set(i);
Z.reset(i+1);
}
else if(index==2){
Z.reset(i);
Z.set(i+1);
}
}
i++;
}
LL ans=0;
rep(i,0,62){
if(Z[i])ans|=1ll<<i;
}
printf("%lld\n",ans);
}
int main(){
int t=rd;
while(t--){
LL n=rd;
find(n);
}
//check::Main();
return 0;
}
/*_________________________________________________________end*/
