TOYOTA SYSTEMS Programming Contest 2021(AtCoder Beginner Contest 228) ABCD

A

题意：

有一个开关，每天s点开，t点关（可能在第2天或第n天），判断x点时开着还是关着。

思路：

按照是否需要隔夜分个类。

#include<bits/stdc++.h>
using namespace std;
int main(){int s, t, x;cin>>s>>t>>x;if(s<t){if(x>=s&&x<t)cout<<"Yes\n";else cout<<"No\n";}else{if(x>=t&&x<s)cout<<"No\n";else cout<<"Yes\n";}return 0;
}

B、

题意：

高桥有n个朋友，每个朋友会向a[i]分享秘密，最开始s知道了秘密，求最后多少人会知道秘密。

思路：

开个set维护知道秘密的人，链式遍历即可。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int main(){int n, s;cin>>n>>s;for(int i = 1; i <= n; i++){cin>>a[i];}set<int>se;int now = s;for(int i = 1; i <= n; i++){if(i!=1 && now==s)break;se.insert(now);now = a[now];}cout<<se.size()<<"\n";return 0;
}

C、

题意：

n个学生参加为期4天的考试，每天300分，已知前三天分数。
求最后一天考完第i个学生能否到达前k名，名次定义为比他分数高的人数+1。

思路：

贪心策略是第i个学生最后一天300分，其他人都0分。
把前三天的总分从大到小排个序，每次去二分第i个学生的排名位置，判断是否在k名内。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int a[maxn], b[maxn];
bool cmp(int x, int y){return x>y;}
int main(){int n, k;  cin>>n>>k;for(int i = 1; i <= n; i++){for(int j = 1; j <= 3; j++){int x;  cin>>x;  a[i] += x;}b[i] = a[i];}sort(b+1,b+n+1,cmp);for(int i = 1; i <= n; i++){int p = lower_bound(b+1,b+n+1,a[i]+300,greater<int>())-b;if(p<=k || a[i]+300>=1200)cout<<"Yes\n";else cout<<"No\n";// if(a[i]+300>=b[k])cout<<"Yes\n";// else cout<<"No\n";}return 0;
}

D、

题意：

初始一个空序列 a[0]到a[2^20]，默认都是-1。
Q个操作，操作1找到a[x]往后第一个不是-1的位置，改成时间戳。操作2询问a[x]的值。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = (1<<20)-1;int fa[maxn+10];
void init(int n){for(int i = 0; i <= n; i++)fa[i]=i;}
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void merge(int x, int y){x=find(x);y=find(y);if(x!=y)fa[x]=y;}
int count(int n){int cnt=0; for(int i = 1; i <= n; i++)if(fa[i]==i)cnt++;return cnt;}LL v[maxn+10];int main(){init(maxn);memset(v,-1,sizeof(v));int q;  cin>>q;while(q--){LL op, x;  cin>>op>>x;if(op==1){int i = find(x&maxn);v[i] = x;fa[i] = find((i+1)&maxn);}else{cout<<v[x&maxn]<<"\n";}}return 0;
}

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TOYOTA SYSTEMS Programming Contest 2021(AtCoder Beginner Contest 228) ABCD

A

B、

C、

D、

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