Codeforces Round 909 (Div. 3)

合集 - cf(14)

1.Codeforces Round 909 (Div. 3)2023-12-04

2.Codeforces Round 910 (Div. 2)2023-12-043.Codeforces Round 912 (Div. 2)2023-12-044.Educational Codeforces Round 158 (Rated for Div. 2)2023-12-045.Codeforces Round 911 (Div. 2)2023-12-046.[Educational Codeforces Round 159 (Rated for Div. 2)](https://codeforces.com/contest/1902)2023-12-047.[Codeforces Round 855 (Div. 3)](https://codeforces.com/contest/1800)2023-12-058.Codeforces Round 913 (Div. 3)2023-12-079.Codeforces Round 904 (Div. 2)2023-12-0910.Codeforces Round 914 (Div. 2)2023-12-1011.Codeforces Round 917 (Div. 2)2024-01-0612.Codeforces [Hello 2024]2024-01-1013.Codeforces Round 919 (Div. 2)2024-01-1414.Codeforces Round 920 (Div. 3)2024-01-16

收起

Codeforces Round 909 (Div. 3)

A

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

int n;
void solve(){
    cin>>n;
    for(int i=1;i<=10;i++){
        if(i&1){
            if(n%3==0) n++;
            else{
                cout<<"First"<<endl;
                return;
            }
        }else{
            if(n%3==1) n--;
            else if(n%3==2) n++;
            else n--;
        }
    }
    cout<<"Second"<<endl;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)solve();
    return 0;
}

B

没想到是这么脑残的暴力，，，写半天。

#include <bits/stdc++.h>
#define int unsigned long long
#define endl '\n';
using namespace std;

const int N = 2e5 + 10;
int n;
unsigned long long pre[N];

void solve(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>pre[i];
        pre[i]+=pre[i-1];
    }
    int ans=0;
    for(int len=1;len<=n/2;len++){
        if(n%len) continue;
        int sum1=0;
        int sum2=0;
        for(int i=len;i<=n;i+=len){
            if(sum1==0) sum1=pre[i]-pre[i-len];
            if(sum2==0) sum2=pre[i]-pre[i-len];
            sum1=max(sum1,pre[i]-pre[i-len]);
            sum2=min(sum2,pre[i]-pre[i-len]);
        }
        ans=max(ans,sum1-sum2);
    }
    cout<<ans<<endl;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)solve();
    return 0;
}

C

很典的贪心。代码写的很丑，写晕了。

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

const int N = 2e5 + 10;
int a[N];
int n;

void solve(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    int ans=a[1];
    int sum=max(0ll,a[1]);
    for(int i=2;i<=n;i++){
        if((a[i]&1)^(a[i-1]&1)){
            sum+=a[i];
            ans=max(ans,sum);
            ans=max(ans,a[i]);
            if(sum<=0) sum=0;
        }else{
            if(a[i]>0){
                sum=a[i];
                ans=max(sum,ans);
            }
            else{
                sum=0;
                ans=max(ans,a[i]);
            }
        }
    }
    cout<<ans<<endl;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)solve();
    return 0;
}

D

如果 2的a[i]次方/a[i]2的a[j]次方/a[j] 的话{i，j}就是符合题意的，两边都取对数就是 a[i]-log2(a[i])a[j]-log2(a[j])

(这种用了log2的代码不知道会不会被卡掉)

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

const int N = 2e5 + 10;
int a[N];
long double b[N];
int n;

void solve(){
    map<long double,int> path;
    int ans=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        long double x;
        x=1.0*a[i]-log2(a[i]);
        ans+=path[x];
        path[x]++;
    }
    cout<<ans<<endl;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)solve();
    return 0;
}

E

可以发现如果出现 a[i]<a[i-1] ，那么i之前的所有数都要进行一次操作。所以找最后一个逆序出现的位置就行了。

#include <bits/stdc++.h>
#define int long long
#define endl '\n';
using namespace std;

const int N = 2e5 + 10;
int a[N];
int n;

void solve(){
    cin>>n;
    deque<int> path;
    int minn=-1;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(minn==-1) minn=a[i];
        minn=min(a[i],minn);
        if(a[i]<a[i-1]) path.push_back(i);
    }
    if(path.empty()){
        cout<<0<<endl;
        return;
    }
    int ans=path.back()-1;
    for(int i=1;i<=ans;i++)
        if(a[i]==minn){
            cout<<-1<<endl;
            return;
        }
    cout<<ans<<endl;
}

signed main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)solve();
    return 0;
}

F

写的时候一直没理解题目，，，其实好像拿一个点出来来回跑就行了，

感觉今天这场的题意都怪怪的

看了下别人的代码好像都是维护两条链的.

大佬代码:

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int long long

void solve() {
    int n, q;cin >> n >> q;
    vector<int> list1,list2;
    for(int i=2;i<=n;i++){
        cout<<i-1<<" "<<i<<"\n";
        list1.push_back(i);
    }
    list2.push_back(2);
    for(int i=1;i<=q;i++){
        int x;cin>>x;
        if(list1.size()==x){
            cout<<"-1 -1 -1\n";
            continue;
        }
        if(list1.size()>x){
            vector<int> st;
            while(list1.size()>x){
                st.push_back(list1.back());
                list1.pop_back();
            }
            cout<<st.back()<<" "<<list1.back()<<" "<<list2.back()<<"\n";
            while(!st.empty()){
                list2.push_back(st.back());
                st.pop_back();
            }
        }else{
            vector<int> st;
            while(list1.size()+st.size()<x){
                st.push_back(list2.back());
                list2.pop_back();
            }
            cout<<st.back()<<" "<<list2.back()<<" "<<list1.back()<<"\n";
            while(!st.empty()){
                list1.push_back(st.back());
                st.pop_back();
            }
        }
    }
}

signed main() {
    ios::sync_with_stdio(false);cin.tie(0);
    int t;cin >> t;
    while (t--) solve();
    return 0;
}