Codeforces Round #807 (Div. 2) && Codeforces Round #809 (Div. 2) A~D
今天开始就搬迁到博客园辣！
先809
A
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n,m;cin>>n>>m;
vector<int>a(n+1);
set<int>s;
for(int i=1;i<=n;i++){
cin>>a[i];
if(a[i] < (m-a[i]+1)) {
if (s.count(a[i])==0)s.insert(a[i]);
else s.insert(m - a[i] + 1);
}else{
if (s.count(m - a[i] + 1)==0)s.insert(m - a[i] + 1);
else s.insert(a[i]);
}
}
for(int i=1;i<=m;i++){
if(s.count(i))cout<<'A';
else cout<<'B';
}
cout<<endl;
}
return ~~(0^_^0);
}
B
不难发现我们可以绕偶数个回来 并且要是连续的数字也可以直接放上去 麻烦的是奇偶都要走一遍
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n;cin>>n;
vector<int>v[n+1];
for(int i=1;i<=n;i++){
int x;cin>>x;
v[x].push_back(i);
}
for(int i=1;i<=n;i++){
if(v[i].size()==0){
cout<<0<<' ';
continue;
}
int cnt=1,flag=0;
for(int j=0;j+1<v[i].size();j++){
if(v[i][j+1]==v[i][j]+1||(v[i][j+1]-v[i][j]-1)%2==0)cnt++;
else if(v[i][j+1]!=v[i][j]+1&&(v[i][j+1]-v[i][j]-1)%2!=0)flag=j;
}
int cnt1=1;
if(flag){
for(int j=flag;j+1<v[i].size();j++){
if(v[i][j+1]==v[i][j]+1||(v[i][j+1]-v[i][j]-1)%2==0)cnt1++;
}
}
cout<<max(cnt1,cnt)<<' ';
}
cout<<endl;
}
return ~~(0^_^0);
}
C
最开始想到dp 但是好像不是max不可做 然后就简单的想了一下最多只能往后挪一次 那维护一个前缀和就可以了
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
int q;
signed main(){
fast
int T;cin>>T;
while(T--){
int n;cin>>n;
vector<int>a(n+1);
int sum=0;
for(int i=1;i<=n;i++){
cin>>a[i];
}
vector<int>w(n+1);
for(int i=2;i<=n-1;i++){
w[i]=max(max(a[i-1],a[i+1])+1-a[i],q);
if(i%2==0)sum+=w[i];
}
if(n%2){
cout<<sum<<endl;
continue;
}
vector<int>f(n+1);
for(int i=n-1;i>=3;i-=2){
if(i!=n-1)f[i]+=f[i+2]+w[i];
else f[i]=w[i];
}
int cnt=0,res=inf;
for(int i=2;i<=n-2;i+=2){
res=min(res,cnt+f[i+1]);
cnt+=w[i];
}
cout<<min(min(cnt,f[3]),res)<<endl;
}
return ~~(0^_^0);
}
D1
此题我们可以暴力枚举每一个最小值 然后二分判断其每组里面离他最近的最大值 一定包含最优解
而最小值的范围是啥呢 就是[a1/k,a1] a1/k 肯定是下界 而我们的上界只能是a1 要是a2 的话a1就没有大于a2的值了
时间复杂度是O（n2）
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n,k;cin>>n>>k;
vector<int>a(n+1);
vector<int>v[n+1];
set<int>s;
for(int i=1;i<=n;i++){
cin>>a[i];
for(int j=k;j>=1;j--){
if(v[i].empty()||v[i].back()!=a[i]/j){
v[i].push_back(a[i]/j);
s.insert(a[i]/j);
}
}
}
int l=a[1]/k,r=a[1],ans=inf;
for(int i=l;i<=r;i++){
int mx=0;
if(s.count(i)==0)continue;
for(int j=1;j<=n;j++){
int t=lower_bound(v[j].begin(),v[j].end(),i)-v[j].begin();
mx=max(mx,v[j][t]-i);
}
ans=min(ans,mx);
}
cout<<ans<<endl;
}
return ~~(0^_^0);
}
807
A
我们构造一个最优序列即可
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n,x;cin>>n>>x;
vector<int>a(n*2);
for(int i=0;i<n*2;i++){
cin>>a[i];
}
sort(a.begin(),a.end());
for(int i=0,j=n;j<a.size();i++,j++){
if(a[j]-a[i]>=x)continue;
else {
no;
goto out;
}
}
yes;
out:1;
}
return ~~(0^_^0);
}
B
不难看出规律就是sum+cnt0
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n;cin>>n;
vector<int>a(n+1);
int flag=0,cnt0=0,sum=0;
for(int i=1;i<=n;i++){
cin>>a[i];
if(i<n)sum+=a[i];
if(a[i]&&i<n)flag=1;
if(flag&&a[i]==0&&i<n)cnt0++;
}
if(sum<cnt0){
cout<<sum+cnt0<<endl;
}else{
cout<<sum+cnt0<<endl;
}
}
return ~~(0^_^0);
}
C
没开ll 爆了两次TLE
我们考虑c<=40 那么我们维护每个区间 以及前缀和 每次查询通过区间和前缀查询
考虑到k=1e18 我们二分查找 时间复杂度O(qclogk)
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n,c,q;cin>>n>>c>>q;
vector<char>a(n+1);
for(int i=1;i<=n;i++)cin>>a[i];
pair<int,int>p[c+1];
vector<int>s(c+1);
s[0]=n;
p[0].first=1,p[0].second=n;
for(int i=1;i<=c;i++){
int l,r;cin>>l>>r;
s[i]=s[i-1]+r-l+1;
p[i].first=l,p[i].second=r;
}
while(q--){
int x;cin>>x;
while(x>n){
int t = lower_bound(s.begin(), s.end(), x) - s.begin();
x -= s[t] - p[t].second;
}
cout<<a[x]<<endl;
}
}
return ~~(0^_^0);
}
D
这道题很一眼吧 我们特判左右两边是否相等 然后色块数量是否相等 求距离也是贪心的来求就好了
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 1<<16;
const int mod = 1e9+7;
#define int long long
#define LL long long
#define endl '\n'
#define Endl '\n'
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
signed main(){
fast
int T;cin>>T;
while(T--){
int n;cin>>n;
string a,b;cin>>a>>b;
a='?'+a+'0',b='?'+b+'0';
int l1,r1,l2,r2,cnta=0,cntb=0;
l1=r1=l2=r2=0;
vector<pair<int,int>>va;
vector<pair<int,int>>vb;
int l,r,flag=0;
for(int i=1;i<=n;i++){
if(a[i]=='1'&&flag==0)l=i,flag=1;
if(a[i]=='1'&&a[i+1]=='0'){
cnta++;
r=i;
flag=0;
va.emplace_back(l,r);
}
}
for(int i=1;i<=n;i++){
if(b[i]=='1'&&!flag)l=i,flag=1;
if(b[i]=='1'&&b[i+1]=='0'){
cntb++;
r=i;
flag=0;
vb.emplace_back(l,r);
}
}
if(a[1]=='1')l1++;
if(a[n]=='1')r1++;
if(b[1]=='1')l2++;
if(b[n]=='1')r2++;
if((l1!=l2)||(r1!=r2)){
cout<<-1<<endl;
}else{
if(cnta==cntb){
int ans=0;
for(int i=0;i<va.size();i++){
ans+=abs(va[i].first-vb[i].first)+abs(va[i].second-vb[i].second);
}
cout<<ans<<endl;
}else cout<<-1<<endl;
}
}
return ~~(0^_^0);
}
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