AtCoder Beginner Contest 338（A~E补题）
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_back
using 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;
}
