Educational Codeforces Round 159 (Rated for Div. 2)

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Educational Codeforces Round 159 (Rated for Div. 2)

A

贪心，看1和3的位置即可

#include <bits/stdc++.h>using namespace std;void solve()
{string s;cin >> s;int a = s.find('1') , b = s.find('3');if(a > b)cout << "31\n";else cout << "13\n";
}int main()
{int T;cin >> T;while(T --){solve();}
}

B

当且仅当0和1相同紧挨着存在时成立

#include <bits/stdc++.h>using namespace std;void solve()
{string a , b;cin >> a;cin >> b;int n = a.size() , c0 = -1 , c1 = -1;for(int i = 0 ; i < n - 1; i ++){if(a[i]=='0'&&a[i+1]=='1'&&b[i]=='0'&&b[i+1]=='1'){cout<<"YES\n";return ;}}cout << "NO\n";
}int main()
{int T;cin >> T;while(T --){solve();}
}

C

令cnt为当前数量，0_{n1一定已排序,0}n2一定未排序

#include <bits/stdc++.h>using namespace std;void solve()
{string s;cin >> s;int cnt = 0 , n1 = 0 , n2 = 0 , n = s.size();bool f = true;//0~n1一定已排序,0~n2一定未排序for(int i = 0 ; i < n ;i ++){if(s[i] == '0'){if(cnt < 2 || cnt == n1)f = false;if(n2 == 0)n2 = cnt;}else if(s[i] == '1'){if(n2 != 0)f = false;n1 = cnt;}else if(s[i] == '+'){cnt ++;}else if(s[i] == '-'){cnt --;if(cnt < n1)n1 = cnt;if(cnt < n2)n2 = 0;}}if(f)cout << "YES\n";else cout << "NO\n";
}int main()
{int T;cin >> T;while(T --){solve();}
}

D

枚举正数和负数的边界

#include<bits/stdc++.h>
using namespace std;
int t,n,a[200005];
int main(){cin>>t;while(t--){cin>>n;for(int i=1;i<=n;i++)cin>>a[i];int ans=INT_MAX,now=0,now1=0;//在0分界for(int i=1;i<n;i++)if(a[i]>=a[i+1])now++;ans=min(ans,now);for(int i=1;i<n;i++){//枚举在i分界if(a[i]>=a[i-1])now1++;if(a[i]>=a[i+1])now--;ans=min(ans,now+now1);}cout<<ans<<"\n";}return 0;
}