Codeforces Round 910 (Div. 2)

A

模拟

#include <bits/stdc++.h>using namespace std;
const int N=1e5+10;
char s[N];void solve(){int n , k , cb = 0 , ans = 0;cin >> n >> k;for(int i = 1 ; i <= n ; i ++){cin >> s[i];if(s[i] == 'B')cb ++;}if(cb == k){cout << "0\n";return ;}else{cout << "1\n";    }while(cb < k){for(int i = 1 ; i <= n ; i ++){if(s[i] == 'A'){cb ++;if(cb == k){cout << i << " B" << endl;return ;}}}}while(cb > k){for(int i = 1 ; i <= n ; i ++){if(s[i] == 'B'){cb --;if(cb == k){cout << i << " A" << endl;return ;}}}     }
}int main()
{int T;cin >> T;while(T --){solve();}return 0;
}

B

最后一个不能分解，依次往前计算

#include <bits/stdc++.h>using namespace std;
const int N = 2e5 + 10;
int a[N];void solve()
{int64_t n, res = 0;cin >> n;for (int i = 1; i <= n; i++)cin >> a[i];for (int i = n - 1; i >= 1; i--){if (a[i] > a[i + 1]){res += ceil(1.0 * a[i] / a[i + 1]) - 1;a[i] = a[i] / ceil(1.0 * a[i] / a[i + 1]);}}cout << res << endl;
}int main()
{int T;cin >> T;while (T--){solve();}return 0;
}

C

一半红一半蓝，当多出奇数时非法

#include <bits/stdc++.h>using namespace std;
const int N = 2e5 + 10;
int a[N];void solve()
{int n , m , k;cin >> n >> m >> k;if(n+m-2>k||(k-(n+m-2))%2){cout << "No\n";return ;}cout<<"YES"<<endl;for(int i=1;i<=n;i++){int t=1;for(int j=1;j<m;j++){if(t){cout<<"R ";}else{cout<<"B ";}t^=1;}cout<<endl;}for(int i=1;i<n;i++){for(int j=1;j<=m;j++){if(j<=2&&i==1){cout<<"B ";}else{if((i+j)%2==0){cout<<"R ";}else{cout<<"B ";}}}cout<<endl;}
}int main()
{int T;cin >> T;while (T--){solve();}return 0;
}

D

将每对a,b抽象为线段
相隔时会增加，找相隔最大的

#include <bits/stdc++.h>using namespace std;
const int N = 2e5 + 10;
int a[N] , b[N];void solve()
{long long ans=0 , n;cin>>n;for(int i=1;i<=n;i++)cin>>a[i];for(int i=1;i<=n;i++)cin>>b[i];for(int i=1;i<=n;i++)ans+=abs(a[i]-b[i]);int maxn=0,minn=1e18;//找到最小的右端点和最大的左端点for(int i=1;i<=n;i++){maxn=max(maxn,min(a[i],b[i]));minn=min(minn,max(a[i],b[i]));}cout<<max(ans,ans+2*(maxn-minn))<<'\n';
}int main()
{int T;cin >> T;while (T--){solve();}return 0;
}

---