A.
模拟
// Problem: A. Everyone Loves to Sleep
// Contest: Codeforces - Codeforces Round 811 (Div. 3)
// URL: https://codeforces.com/group/RAx5fiilkP/contest/1714/problem/A
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
struct node{int h,m;
}a[N];
bool cmp(node a,node b){if(a.h!=b.h){return a.h<b.h;}return a.m<b.m;
}
void Lan(){int n,h,m;cin>>n>>h>>m;for(int i=1;i<=n;i++){cin>>a[i].h>>a[i].m;if(a[i].h<h || (a[i].h==h && a[i].m<m)){a[i].h+=24;}}sort(a+1,a+1+n,cmp);int ans=h*60+m;int res=a[1].h*60+a[1].m;if(ans==res){cout<<0<<" "<<0<<'\n';return;}res-=ans;if(res%60==0){cout<<res/60<<" "<<0<<'\n';}else{cout<<res/60<<" "<<res%60<<'\n';}}
int main() {ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int q;cin>>q;while (q--) {Lan();}return 0;
}B.
从后往前遍历一遍即可
// Problem: B. Remove Prefix
// Contest: Codeforces - Codeforces Round 811 (Div. 3)
// URL: https://codeforces.com/group/RAx5fiilkP/contest/1714/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
int a[N],vis[N];
void Lan(){int n;cin>>n;int mx=0;for(int i=1;i<=n;i++){cin>>a[i];mx=max(mx,a[i]);}for(int i=1;i<=(mx+1);i++){vis[i]=0;}for(int i=n;i>=1;i--){if(!vis[a[i]]){vis[a[i]]=1;}else{cout<<i<<'\n';return;}}cout<<0<<'\n';
}
int main() {ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int q;cin>>q;while (q--) {Lan();}return 0;
}C.
简单贪心
// Problem: C. Minimum Varied Number
// Contest: Codeforces - Codeforces Round 811 (Div. 3)
// URL: https://codeforces.com/group/RAx5fiilkP/contest/1714/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
int a[N];
void Lan(){int s;cin>>s;vector<int> v;int num=9;while(s){if(s>num){s-=num;v.push_back(num);num--;}else{v.push_back(s);break;}}reverse(v.begin(),v.end());for(auto &i:v){cout<<i;}cout<<'\n';
}
int main() {ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int q;cin>>q;while (q--) {Lan();}return 0;
}D.
1.
重点
学到了区间覆盖
先暴力枚举有几个可以覆盖
然后贪心
找最长的线段去覆盖
认定左边都是被上一个覆盖过了，所以只要枚举右边线段即可
因为写的是while所以我玄学了加了一个计时器如果超过1000就说明不存在方案即可
debug...
2.
dp
// Problem: D. Color with Occurrences
// Contest: Codeforces - Codeforces Round 811 (Div. 3)
// URL: https://codeforces.com/group/RAx5fiilkP/contest/1714/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 1e5 + 9;
int dp[N],pre[N];
pair<int,int>a[N];
string ss[N];
void Lan(){string t;cin>>t;int m=t.length();t=" "+t;int n;cin>>n;for(int i=1;i<=n;i++){cin>>ss[i];}memset(dp,0x3f,sizeof(dp));dp[0]=0;int mx=dp[1];for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){if(i>=ss[j].length()){if(t.substr(i-ss[j].length()+1,ss[j].length())==ss[j]){for(int l=(int)(i-ss[j].length());l<i;l++){if(dp[l]+1<dp[i]){dp[i]=dp[l]+1;pre[i]=l;a[i]={i-ss[j].length()+1,j};}}}}}}if(dp[m]==mx){cout<<-1<<'\n';}else{cout<<dp[m]<<'\n';int now=m;while(now){cout<<a[now].second<<" "<<a[now].first<<'\n';now=pre[now];}}}
int main() {ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int q;cin>>q;while (q--) {Lan();}return 0;
}E.
找规律举几个数字
// Problem: E. Add Modulo 10
// Contest: Codeforces - Codeforces Round 811 (Div. 3)
// URL: https://codeforces.com/group/RAx5fiilkP/contest/1714/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
int a[N];
//发现有5和没5倍数在一起就不可能
//其他就抽成只有2，然后这样变化就+20，mod20看最后数字是不是一样即可
int work(int x){while(x%10!=2 && x%10!=0){x=x+x%10;}return x;
}
void Lan(){int n;cin>>n;for(int i=1;i<=n;i++){cin>>a[i];}int cnt=0;for(int i=1;i<=n;i++){if(a[i]%5==0){cnt++;}}if(cnt>0 && cnt<n){cout<<"No"<<'\n';return;}for(int i=1;i<=n;i++){a[i]=work(a[i]);}if(!cnt){for(int i=1;i<=n;i++){a[i]%=20;}}for(int i=1;i<=n-1;i++){if(a[i]!=a[i+1]){cout<<"No"<<'\n';return;}}cout<<"Yes"<<'\n';
}
int main() {ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);int q;cin>>q;while (q--) {Lan();}return 0;
}
)
