D
这是一个环 每个点跑一遍lis找到最大 n-max就是答案
#include <bits/stdc++.h> #define ll long long using namespace std; int const N=505; int n,dp[N],a[N],ans,ins; int main() { ios::sync_with_stdio(false); cin>>n; for(int i=1;i<=n;++i) { cin>>a[i]; } for(int i = 0;i < n;i ++) { memset(dp, 0, sizeof(dp)); for(int j = 1;j <= n;j ++) { dp[j] = 1; for(int k = 1;k < j;k ++) { if(a[(i + j) % n + 1] > a[(i + k) % n + 1]) dp[j] = max(dp[j], dp[k] + 1); } ins = max(ins, dp[j]); } ans = max(ans, ins); } cout<<n-ans<<endl; return 0; }
E
这是我们队的黄大佬用java写的 看伪代码找最小循环节
import java.util.*; import java.math.BigInteger; public class Main { public static void main(String[] args) { Scanner s = new Scanner(System.in); int n = s.nextInt(); int[] a = new int[n+1]; int[] vis = new int[n+1]; BigInteger c = new BigInteger("1"); BigInteger l = new BigInteger("1"); for(int i=1;i<=n;++i)a[i] = s.nextInt(); for(int i=1;i<=n;++i) { if(vis[i] != 0)continue; int x = a[i]; BigInteger tot = new BigInteger("0"); do { vis[x] = tot.intValue(); x = a[x]; tot = tot.add(l); } while(vis[x] == 0&&x != a[i]); if(vis[x] != 0)tot.add(new BigInteger(String.valueOf(vis[x]))); c = c.multiply(tot).divide(c.gcd(tot)); } System.out.println(c); } }
F
简单的模拟 看懂题意就行
#include<bits/stdc++.h> using namespace std; #define int long long inline int nextInt() { char c = getchar(); int x = 0,fh = 0; while(c < '0' || c > '9'){fh |= c == '-';c = getchar();} while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();} return fh?-x:x; } signed main() { int n = nextInt(); int a[101],maxa = 0; for(int i=1;i<=n;++i)maxa = max(maxa,a[i] = nextInt()); for(int i=1;i<=n;++i) { int t = ceil(50.0*a[i]/maxa); cout<<'+';for(int i=1;i<=t;++i)cout<<"-";cout<<'+'<<endl; cout<<'|'; if(a[i] == maxa) { for(int i=1;i<=t-1;++i)cout<<' '; cout<<'*'; } else for(int i=1;i<=t;++i)cout<<' '; cout<<'|'<<a[i]<<endl; cout<<'+';for(int i=1;i<=t;++i)cout<<"-";cout<<'+'<<endl; } return 0; }
I
在无穷大的平面上 每个特殊点必定会对4个基地点产生贡献 有两种特殊点
所以答案是2/3
#include <bits/stdc++.h> #define ll long long using namespace std; int n; int main() { ios::sync_with_stdio(false); double ans=2.0/3; printf("%.6f",ans); return 0; }
具体看代码理解