A. Twice
看相同数字出现次数即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now << 1
#define rs now << 1 | 1
#define endl "\n"
#define lowbit(x) ((x) & (-x))
typedef long long ll;
const int N = 1e5 + 7, mod = 1e9 + 7;
int n;
int a[N];
void solve() {
map<int,int>vis;
int ans=0;
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
vis[a[i]]++;
if(vis[a[i]]>=2)ans++,vis[a[i]]-=2;
}
cout<<ans<<endl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
// freopen("..//..//in_out//in.txt", "r", stdin);
// freopen("..//..//in_out//out.txt", "w", stdout);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
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显然是找两个数相乘等于 $n-2$ 即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now << 1
#define rs now << 1 | 1
#define endl "\n"
#define lowbit(x) ((x) & (-x))
typedef long long ll;
const int N = 2e5 + 7, mod = 1e9 + 7;
int n;
int a[N];
void solve() {
cin>>n;
map<int,int>vis;
for(int i=1;i<=n;i++){
cin>>a[i];
vis[a[i]]++;
}
int tmp=n-2;
for(int i=1;i<=n;i++){
int t=tmp/a[i];
if(t*a[i]==tmp){
if(t==a[i]&&vis[a[i]]>=2){
cout<<t<<" "<<t<<endl;
return;
}
else if(t=a[i]&&vis[t]){
cout<<a[i]<<" "<<t<<endl;
return;
}
}
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
// freopen("..//..//in_out//in.txt", "r", stdin);
// freopen("..//..//in_out//out.txt", "w", stdout);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
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C. Superultra’s Favorite Permutation
可以发现，小于 $5$ 的都不行，然后先按 ${1,3,5,4,2}$ 排，奇数插左边，偶数插右边即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now << 1
#define rs now << 1 | 1
#define endl "\n"
#define lowbit(x) ((x) & (-x))
typedef long long ll;
const int N = 2e5 + 7, mod = 1e9 + 7;
deque<int>q;
void solve() {
int n;
q.clear();
cin>>n;
if(n<=4){
cout<<"-1\n";
return;
}
q.push_back(1);q.push_back(3);q.push_back(5);q.push_back(4);q.push_back(2);
for(int i=6;i<=n;i++){
if(i&1)q.push_front(i);
else q.push_back(i);
}
for(auto x:q)cout<<x<<" ";
cout<<endl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
// freopen("..//..//in_out//in.txt", "r", stdin);
// freopen("..//..//in_out//out.txt", "w", stdout);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
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Sharky Surfing
显然，当前所处位置左侧的加速点都能吃到，只需要保证当前能量足够跳过当前右侧的障碍即可，模拟一遍即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now << 1
#define rs now << 1 | 1
#define endl "\n"
#define lowbit(x) ((x) & (-x))
typedef long long ll;
const int N = 2e5 + 7, mod = 1e9 + 7;
int n,m,L;
void solve() {
cin>>n>>m>>L;
deque<pair<int,int>>block,power;
for(int i=1;i<=n;i++){
int a,b;
cin>>a>>b;
block.push_back({a,b});
}
for(int i=1;i<=m;i++){
int a,b;
cin>>a>>b;
power.push_back({a,b});
}
int pos=1,now=1,ans=0;
multiset<int>s;
while(block.empty()){
while(block.empty() && pos >= block.front().second) block.pop_front();
if (block.empty()) break;
pos=block.front().first;
while(power.size()){
if(power.front().first<pos){
s.insert(power.front().second);
power.pop_front();
}
else break;
}
int len=block.front().second-block.front().first+1;
while(now<=len&&s.size()){
ans++;
now+=*(--s.end());
s.erase((--s.end()));
}
if(now<=len){
cout<<"-1\n";
return;
}
pos+=now;
}
cout<<ans<<endl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
// freopen("..//..//in_out//in.txt", "r", stdin);
// freopen("..//..//in_out//out.txt", "w", stdout);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
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E. Kachina’s Favorite Binary String
题看半天才看懂
$f(l,r)$ 是这个子串中存在 $01$ 子序列的数量
IMPOSSIBLE 的情况就是 $f(1,n)$ 等于 $0$，这时候可能全 $1$ 或全 $0$
有 $n$ 次询问机会，那就可以固定左端点为 $1$，右端点 $2~n$，将对应结果存放起来
因为要找的是 $01$，然后起始连续的 $1$ 都不会对后面造成影响，即我们需要找到第一次 $0$ 出现的位置
对每个位置询问的结果和他前方的比较，如果当前位置比前方大，那么当前位置一定是 $1$，才能使 $f$ 的值增大，反之就是 $0$
接着找到第一次出现 $0$ 的位置，此位置前方有一个片段都是 $1$，片段从 $1$ 到 $i-1-cnt[i]$，这样保证前方的 $01$ 刚好符合题意，前方符合后，后方也会符合
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#include <bits/stdc++.h>
using namespace std;
#define ls now<<1
#define rs now<<1|1
// #define endl "\n"
#define lowbit(x) ((x)&(-x))
typedef long long ll;
const int N=1e4+7, mod=1e9+7;
int n;
int a[N],b[N];
void solve(){
cin>>n;
for(int i=0;i<=n;i++)b[i]=0;
for(int i=2;i<=n;i++){
cout<<"? 1 "<<i<<"\n";
cin>>a[i];
}
if(a[n]==0){
cout<<" IMPOSSIBLE\n";
cout<<endl;
return;
}
b[1]=0;
for(int i=2;i<=n;i++){
if(a[i]>a[i-1])b[i]=1;
}
for(int i=2;i<=n;i++){
if(b[i]){
for(int j=1;j<=i-1-a[i];j++)b[j]=1;
}
}
cout<<" ";
for(int i=1;i<=n;i++)cout<<b[i];
cout<<endl;
}
int main(){
// ios::sync_with_stdio(0);
// cin.tie(0);cout.tie(0);
int t=1;
cin>>t;
while(t--)solve();
return 0;
}
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