A

签到

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;const int N=5e5+10;void solve()
{string str;cin>>str;bool st=true;rep(i,0,str.size()-1){if(!i)	{if(str[i]>='A'&&str[i]<='Z')	continue;else{cout<<"No"<<endl;return;}}if(str[i]>='a'&&str[i]<='z')	continue;else{cout<<"No"<<endl;return;}}cout<<"Yes"<<endl;
}int main()
{IOS	
//  	freopen("1.in", "r", stdin);int t;
//	cin>>t;
//	while(t--)solve();return 0;
}

B

签到题

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;const int N=5e5+10;void solve()
{string str;cin>>str;map<char,int>cnt;for(auto c:str)	cnt[c]++;char c;int k=0;for(auto it:cnt){if(it.y>k)	{k=it.y;c=it.x;}}cout<<c<<endl;
}int main()
{IOS	
//  	freopen("1.in", "r", stdin);int t;
//	cin>>t;
//	while(t--)solve();return 0;
}

C题

观察到数据范围n只有10，想到去枚举a的数量然后去看b最多能做多少。

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;const int N=20;
int c[N],a[N],b[N],n;bool check1(int x)
{rep(i,1,n)	if(x*a[i]>c[i])	return false;return true;
}void solve()
{cin>>n;int sa=0,sb=0,sc=0;rep(i,1,n)	cin>>c[i],sc+=c[i];rep(i,1,n)	cin>>a[i],sa+=a[i];rep(i,1,n)	cin>>b[i],sb+=b[i];int ans=0;rep(gg,0,1e6){if(!check1(gg))	break;int kk=1e6; rep(i,1,n){if(b[i]==0)	continue;int xx=c[i]-a[i]*gg;xx/=b[i];kk=min(kk,xx);} ans=max(ans,gg+kk);}	cout<<ans<<endl;
}int main()
{IOS	
//  	freopen("1.in", "r", stdin);
//  	int t;
//	cin>>t;
//	while(t--)solve();return 0;
}

D题

也很巧妙，贡献法的思想。
对于每一个路径计算他对不同断桥产生的贡献。

#include <bits/stdc++.h> 
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
using namespace std;const int N=1e6+10;ll n,m,x[N],cf[N];ll dist(ll x,ll y)
{if(x<=y)	return y-x;else	return n-(x-y);
}void add(ll x,ll y,ll d)
{if(x<=y){cf[x]+=d;cf[y]-=d;}else{cf[x]+=d;cf[n+1]-=d;cf[1]+=d;cf[y]-=d;}
}void solve()
{cin>>n>>m;rep(i,1,m)	cin>>x[i];rep(i,1,m-1){add(x[i],x[i+1],dist(x[i+1],x[i]));add(x[i+1],x[i],dist(x[i],x[i+1]));}ll ans=1e18;rep(i,1,n)	cf[i]+=cf[i-1];rep(i,1,n)	ans=min(ans,cf[i]);cout<<ans<<endl;
}int main()
{IOS	
//  	freopen("1.in", "r", stdin);int t;
//	cin>>t;
//	while(t--)solve();return 0;
}

E题

#include <bits/stdc++.h> 
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define pii pair<int, int>
#define pll pair<long long, long long>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
#define x first
#define y second
#define pb push_backusing namespace std;
const int N=4e5+10;
int cf[N],x[N],y[N];void solve()
{int n;cin>>n;rep(i,1,n){cin>>x[i]>>y[i];if(x[i]>y[i])	swap(x[i],y[i]);cf[x[i]]+=1;cf[y[i]+1]-=1;	}rep(i,1,2*n)	cf[i]+=cf[i-1];rep(i,1,n)	{if(cf[x[i]]!=cf[y[i]]){cout<<"Yes"<<endl;return;}}cout<<"No"<<endl;	
}
signed main()
{IOS	
//  	freopen("1.in", "r", stdin);
//  	int _;
//	cin>>_;
//	while(_--)solve();return 0;
}