基本全是水题

第一题水,不过有hack点,先增后不变再减

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f; int main()
{
ios::sync_with_stdio(false);
cin.tie();
int n;
cin>>n;
bool f=;
int last,bh=;
for(int i=;i<n;i++)
{
int a;
cin>>a;
if(i==)last=a;
else
{
if(last<a)
{
if(bh==)last=a;
else if(bh==)f=;
else f=;
}
else if(last==a)
{
if(bh==)bh=,last=a;
else if(bh==)last=a;
else f=;
}
else if(last>a)
{
if(bh==)bh=,last=a;
else if(bh==)bh=,last=a;
else last=a;
}
}
}
if(f)cout<<"NO"<<endl;
else cout<<"YES"<<endl;
return ;
}

A

第二题更水,6分钟1a,maphash一下就行了

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f; map<char,char>m;
int main()
{
ios::sync_with_stdio(false);
cin.tie();
string s,p,t;
cin>>s>>p>>t;
for(int i=;i<s.size();i++)
{
m[s[i]]=p[i];
}
for(int i=;i<t.size();i++)
{
if('a'<=t[i]&&t[i]<='z')cout<<m[t[i]];
else if('A'<=t[i]&&t[i]<='Z')cout<<(char)(m[t[i]-'A'+'a']+'A'-'a');
else cout<<t[i];
}
cout<<endl;
return ;
}

B

第三题想法题,先sort,然后找间距一一对应起来就可以 了(刚开始想的是求出值来,太蠢了)

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f; ll a[N],b[N];
ll sum[N];
set<ll>ans;
int main()
{
ios::sync_with_stdio(false);
cin.tie();
ll k,n;
cin>>k>>n;
for(ll i=;i<k;i++)
{
cin>>a[i];
if(i==)sum[i]=a[i];
else sum[i]=sum[i-]+a[i];
}
for(ll i=;i<n;i++)cin>>b[i];
sort(b,b+n);
sort(sum,sum+k);
int p=b[];
for(int i=;i<n;i++)b[i]-=p;
ll i=;
for(int i=;i<k;i++)
{
int res=,p=;
for(int j=i;j<k;j++)
{
if(sum[j]-sum[i]==b[p])
p++;
if(p>=n)
{
ans.insert(sum[i]);
break;
}
}
}
cout<<ans.size()<<endl;
return ;
}

C

第四题,给一堆人,和一堆钥匙的坐标,每个人拿钥匙去办公室,求最小时间,sort一下,然后二分最小时间,每次判断的时候还是枚举两个数组进行匹配

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f; ll a[N],b[N],p,n,k;
bool ok(ll i,ll j,ll x)
{
return abs(a[i]-b[j])+abs(b[j]-p)<=x;
}
bool check(ll x)
{
ll j=;
for(int i=;i<n;i++)
{
while(!ok(i,j,x)){
j++;
}
if(j>=k)return ;
j++;
}
return ;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie();
cin>>n>>k>>p;
for(ll i=;i<n;i++)cin>>a[i];
for(ll i=;i<k;i++)cin>>b[i];
sort(a,a+n);
sort(b,b+k);
ll l=,r=2e9+;
for(int i=;i<;i++)
{
ll m=(l+r)/;
if(check(m))r=m;
else l=m;
}
while(check(l))l--;
cout<<l+<<endl;
return ;
}

D

第5.6先留坑

话说这场只a了两题,还好比较稳,没被hack,还涨了85分>.<

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