A.Little Artem
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e5+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
for(int i=1;i<n;i++,cout<<endl)
for(int j=1;j<=m;j++)
if(j==1)cout<<'B';
else cout<<'W';
for(int i=1;i<=m;i++)
cout<<'B';
cout<<endl;
}
return 0;
}
B.Kind Anton
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e5+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int a[maxn],b[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
cin>>t;
while(t--){
int n;
cin>>n;
map<int,int> m;
for(int i=1;i<=n;i++)cin>>a[i];
for(int i=1;i<=n;i++)cin>>b[i];
bool f=0;
for(int i=1;i<=n;i++){
if(a[i]==b[i]){m[a[i]]++;continue;}
int d=b[i]-a[i];
if(d>0&&m[1]){m[a[i]]++;continue;}
if(d<0&&m[-1]){m[a[i]]++;continue;}
f=1;m[a[i]]++;
}
if(f)cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
return 0;
}
C.Eugene and an array
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=2e5+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
ll a[maxn],sum[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i],sum[i]=sum[i-1]+a[i];
map<ll,int> m;
int l=0;
ll ans=0;
for(ll i=1;i<=n;i++){
if(!m[sum[i]]&&sum[i]&&a[i])ans+=(i-l);
else if(!a[i])l=i;
else l=max(l,m[sum[i]]+1),ans+=(i-l);
m[sum[i]]=i;
}
cout<<ans;
return 0;
}
D.Challenges in school №41
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=3e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
vector<int> v[100010];
vector<int> ans[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n,k;
cin>>n>>k;
string s;
cin>>s;
int cnt=0,tot=0;
while(1){
cnt++;
bool f=0;
for(int i=0;i<n;i++)
if(s[i]=='R'&&s[i+1]=='L')
v[cnt].pb(i+1),swap(s[i],s[i+1]),f=1,tot++,i++;
if(!f)break;
if(cnt>k){cout<<-1;return 0;}
}
if(k>tot){cout<<-1;return 0;}
int c=0;
for(int i=1;i<cnt;i++){
for(auto j:v[i]){
if(cnt-i<k)ans[c++].pb(j),k--;
else ans[c].pb(j);
}
if(cnt-i>=k&&!ans[c].empty())c++,k--;
}
for(int i=0;i<c;i++){
cout<<(int)ans[i].size()<<' ';
for(auto j:ans[i])cout<<j<<' ';
cout<<endl;
}
return 0;
}
F.Kate and imperfection
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=5e5+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int a[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
cin>>n;
for(int i=2;i<=n;i++)
if(!a[i])
for(int j=i;j<=n;j+=i)
if(!a[j])a[j]=j/i;
a[1]=1;
vector<int> v;
for(int i=1;i<=n;i++)
if(a[i])v.pb(a[i]);
sort(all(v));
for(int i=1;i<n;i++)
cout<<v[i]<<' ';
return 0;
}