Codeforces Round 910 (Div. 2)

A. Milica and String

wa麻了，，，不知道自己在干什么

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

void solve(){
    int k,n;
    string s;
    cin>>n>>k>>s;
    int cnt=0;
    s = " " + s + " ";
    for(int i=n+1;i;i--) if(s[i]=='B') cnt++;
    if(cnt==k) cout<<0<<endl;
    else if(cnt>k){
        int cnt1=0;
        cout<<1<<endl;
        for(int i=n+1;i>0;i--){
            if(s[i]=='B') cnt1++;
            if(cnt1==k){
                 cout<<i-1<<" A"<<endl;
                 return;
            }
        }
    }else{
        int cnt1=0;
        cout<<1<<endl;
        for(int i=1;i<=n;i++){
            if(s[i]=='A') cnt1++;
            if(cnt1==k-cnt){
                cout<<i<<" B"<<endl;
                return;
            }
        }
    }
   
}

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

B. Milena and Admirer

今天不宜写题

#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;
    int ans=0;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=n-1;i>0;i--){
        if(a[i]<=a[i+1]) continue;
        int x=(a[i] + a[i+1] - 1)/a[i+1];
        a[i]/=x;
        ans+=x-1;
    }
    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. Colorful Grid

写着有点恶心，，，自己的码太丑了贴个佬的码

只要在起点处加个环，终点加个U型就行了

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
using LL = long long;

int main(){

    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int T;
    cin >> T;
    while(T--){
        int n, m, k;
        cin >> n >> m >> k;
        if (k < n + m - 2){
            cout << "NO" << '\n';
            continue;
        }
        if (k % 2 != (n + m - 2) % 2){
            cout << "NO" << '\n';
            continue;
        }
        vector<vector<int> > a1(n, vector<int>(m - 1)), a2(n - 1, vector<int>(m));
        for(int i = 0; i < n - 1; i++){
            if (i % 2 == 0){
                a2[i][0] = 1;
            }
        }
        for(int i = 0; i < m - 1; i++){
            if ((n + i - 1) % 2 == 0){
                a1[n - 1][i] = 1;
            }
        }
        if (k % 4 != (n + m - 2) % 4){
            a1[0][0] = a1[1][0] = 1;
        }
        a2[n - 2][m - 1] = a1[n - 1][m - 2] ^ 1;
        a1[n - 2][m - 2] = a1[n - 1][m - 2];
        a2[n - 2][m - 2] = a2[n - 2][m - 1];
        cout << "YES" << '\n';
        for(int i = 0; i < a1.size(); i++){
            for(auto x : a1[i]){
                cout << (x ? "R" : "B") << ' ';
            }
            cout << '\n';
        }
        for(int i = 0; i < a2.size(); i++){
            for(auto x : a2[i]){
                cout << (x ? "R" : "B") << ' ';
            }
            cout << '\n';
        }
    }

}

D

将每个集合当作一个线段。如果两个线段相交，对他们进行交换结果不变。如果两个线段不相交，假设[l1,r1],[l2,r2],l1<r1<l2<r2，对他们交换结果增大2*(l2-r1),因为只交换一次，所以我们要找到最大的那个左端点和最小的那个右端点。

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

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

void solve(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=n;i++) cin>>b[i];
    int sum=0;
    int maxl=0;
    int minr=1e9;
    for(int i=1;i<=n;i++){
        sum+=abs(a[i]-b[i]);
        maxl=max(maxl,min(a[i],b[i]));
        minr=min(minr,max(a[i],b[i]));
    }
    if(maxl>=minr)sum+=2*(maxl-minr);
    cout<<sum<<endl;
}

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