Codeforces Round #633(Div.2)
\(A.Filling\ Diamonds\)
答案就是构成的六边形数量+1
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
void solve(){
LL n; cin >> n;
cout << n << endl;
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}
\(B.Sorted\ Adjacent\ Differences\)
考虑排完序之后从中间开始一左一右选择即可
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 2e5+7;
int n,A[MAXN];
void solve(){
cin >> n;
for(int i = 1; i <= n; i++) cin >> A[i];
sort(A+1,A+1+n);
int mid = n / 2 + 1;
int lp = mid - 1, rp = mid + 1;
cout << A[mid] << ' ';
while(lp!=0 or rp!=n+1){
if(lp!=0) cout << A[lp--] << ' ';
if(rp!=n+1) cout << A[rp++] << ' ';
}
cout << endl;
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}
\(C.Powered\ Addition\)
考虑对于每个数，需要加的最小值为其之前的最大值和这个值的差值，然后找到这个差值的最大值找到其二进制最高位即可
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
using LL = int_fast64_t;
int n;
LL A[MAXN];
void solve(){
cin >> n;
for(int i = 1; i <= n; i++) cin >> A[i];
LL minn = 1e10;
LL delta = 0;
for(int i = n; i >= 1; i--){
delta = max(delta,A[i]-minn);
minn = min(A[i],minn);
}
if(!delta) cout << 0 << endl;
else{
while(delta!=(delta&-delta)) delta -= (delta&-delta);
cout << __builtin_ctz(delta)+1 << endl;
}
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}
\(D.Edge\ Weight\ Assignment\)
考虑最少不同的数，各个叶子节点到根的距离的奇偶性如果一样，那么所有边权都可以相等，否则的话用\(1,2,3\)三个数来赋值边权必然可以满足条件
对于最多的不同的数，考虑树形\(DP\)，\(dp[i]\)表示以\(i\)为根的子树所用的最多的不同边权，以非叶节点为根进行\(dfs\)，考虑转移，对于\(u\)点的所有儿子，如果存在叶子节点的话，\(dp[u]+=1\)，叶子节点连上来的边权必然相等，对于儿子中的非叶子节点\(v\)，转移为\(dp[u] += dp[v] + 1\)，可以假设叶子节点走到儿子节点的权值异或和为\(x\)，那么不同儿子子树中的叶子节点走到该儿子节点的异或和可以不同，所以从儿子连到当前点的边权也可以不一样。
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
using LL = int_fast64_t;
int n,depth[MAXN],f[MAXN];
vector<int> G[MAXN];
bool odd = false, even = false;
void dfs(int u, int par){
depth[u] = depth[par] + 1;
bool zero = false;
if(par and G[u].size()==1){
if(depth[u]&1) odd = true;
else even = true;
}
for(int v : G[u]){
if(v==par) continue;
dfs(v,u);
if(f[v]==0) zero = true;
else f[u] += f[v] + 1;
}
if(zero) f[u]++;
}
int main(){
____();
cin >> n;
for(int i = 1; i < n; i++){
int u, v;
cin >> u >> v;
G[u].emplace_back(v);
G[v].emplace_back(u);
}
int root = 1;
for(int i = 1; i <= n; i++) if(G[i].size()>1) root = i;
dfs(root,0);
cout << ((odd and even)?3:1) << ' ' << f[root] << endl;
return 0;
}
\(C.Perfect\ Triples\)
找规律题
对着表找了半天规律没写出来，我菜死了
用四进制表示每一位，可以打表出来
001 002 003
//1行
010 020 030
011 022 033
012 023 031
013 021 032
//1<<2行
100 200 300
101 202 303
102 203 301
103 201 302
110 220 330
111 222 333
112 223 331
113 221 332
... ... ...
//1<<4行
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
void solve(){
LL n; cin >> n;
LL m = (n - 1) / 3 + 1, md = n % 3 ? n % 3 : 3, ret = 0, bit = 1, num = 0, a[4];
if(md==1) a[0] = 3, a[1] = 0, a[2] = 1, a[3] = 2;
else if(md==2) a[0] = 1, a[1] = 0, a[2] = 2, a[3] = 3;
else a[0] = 2, a[1] = 0, a[2] = 3, a[3] = 1;
while(true){
if(bit>=m){
ret += (md<<num);
break;
}
m -= bit;
ret += (a[((m-1) / bit + 1) % 4]<<num);
bit <<= 2, num += 2;
}
cout << ret << endl;
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}