Educational Codeforces Round 170 (Rated for Div. 2)
比赛链接：https://codeforces.com/contest/2025
A. Two Screens
看前面最长相同前缀，如果有就前缀长+1+剩余长度，否则直接两个字符串长度相加
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll a[250000];
int main()
{
ll t;
cin>>t;
while(t--)
{
string a,b;
cin>>a>>b;
ll ans=0;
ll wz=-1;
for(ll i=0;i<=min(a.size()-1,b.size()-1);i++)
{
if(a[i]==b[i])
{
ans++;
wz=i;
}
else
{
break;
}
}
if(wz!=-1)
{
ans++;
}
//cout<<ans<<endl;
ans+=(ll)a.size()-1-wz;
ans+=(ll)b.size()-1-wz;
cout<<ans<<endl;
}
}
B. Binomial Coefficients, Kind Of
手推算得到，k=1或者k=0答案为1，其余则是\(2^k\)
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll a[250000];
ll b[250000];
int main()
{
ll n;
cin>>n;
for(ll i=1;i<=n;i++)cin>>a[i];
for(ll i=1;i<=n;i++)cin>>b[i];
for(ll i=1;i<=n;i++)
{
if(a[i]==b[i]||b[i]==0)
{
cout<<1<<endl;
}
else
{
cout<<ksm(2,b[i])%mod<<endl;
}
}
}
C. New Game
本质就是维护队列，先离散化每个数，然后维护两个条件下的队列即可
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll a[250000];
set<ll>q;
map<ll,ll>f;
int main()
{
ll t;
cin>>t;
while(t--)
{
q.clear();
f.clear();
ll n,k;
cin>>n>>k;
for(ll i=1;i<=n;i++)
{
cin>>a[i];
q.insert(a[i]);
f[a[i]]++;
}
deque<ll>op;
ll ans=0;
ll cnt=0;
for(auto j:q)
{
if(op.empty())
{
op.push_back(j);
cnt+=f[j];
ans=max(ans,cnt);
}
else
{
if(j==op.back()+1)
{
cnt+=f[j];
op.push_back(j);
while((ll)op.size()>k)
{
cnt-=f[op.front()];
op.pop_front();
}
ans=max(ans,cnt);
}
else
{
op.clear();
op.push_back(j);
cnt=f[j];
ans=max(ans,cnt);
}
}
}
cout<<ans<<endl;
}
}
D. Attribute Checks
用一维数组差分每次选择后的可能变化，然后每次到0就结算之前的差分数组，以智力为什么时进行dp，最大值可以选择前面的或者现在的.
参考jly代码
#include<iostream>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<unordered_set>
#include<stack>
#include<queue>
#include<algorithm>
#include<time.h>
#include<random>
#include<string>
#include<string.h>
#include<cctype>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
const ll mod = 1e9 + 7;
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
ll ksm(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans = ans%mod*(x%mod)%mod;
x = x%mod*(x%mod)%mod;
y >>= 1;
}
return ans%mod%mod;
}
ll dp[6005];
ll a[3500000];
ll f[6005];
int main()
{
fio();
ll n,m;
cin>>n>>m;
ll cnt=0;
for(ll i=1;i<=n;i++)
{
cin>>a[i];
if(a[i]==0)
{
for(ll j=0;j<=m;j++)
{
if(j>0)
{
dp[j]+=dp[j-1];
dp[j-1]=0;
}
f[j]+=dp[j];
}
for(ll j=m;j>=1;j--)
{
f[j]=max(f[j],f[j-1]);
}
dp[m]=0;
cnt++;
continue;
}
if(a[i]>0)
{
if(cnt>=a[i])
{
dp[a[i]]++;
dp[m+1]--;
}
}
else
{
ll u=abs(a[i]);
if(cnt>=u)
{
dp[0]++;
dp[cnt-u+1]--;
}
}
}
for(ll j=0;j<=m;j++)
{
if(j>0)
{
dp[j]+=dp[j-1];
dp[j-1]=0;
}
f[j]+=dp[j];
}
for(ll j=m;j>=1;j--)
{
f[j]=max(f[j],f[j-1]);
}
ll ans=0;
for(ll j=0;j<=m;j++)
{
ans=max(ans,f[j]);
}
cout<<ans<<endl;
}
