1479A - Searching Local Minimum 交互，二分，2019 ccpc 哈尔滨 E 拓扑排序

1479A - Searching Local Minimum 交互，二分

找一个区间[l,r]始终满足a[l+1]>a[l]&&a[r]<a[r+1],然后不断缩小区间当l==r时就找到了答案，二分出mid，然后询问mid和mid+1,如果a[mid]>a[mid+1]则区间[mid+1,r]还是满足条件的,l=mid+1;如果a[mid]<a[mid+1]则区间[l,mid]是满足条件的，r=mid;

#700 (Div. 2)C. Searching Local Minimum（二分的神奇用法+交互）_lwz_159的博客-CSDN博客

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define lowbit(x) ((x)&(-x))
#define int long long
#define pause system("pause")
const int mod=1e9+7;
const int inf=1e18;
const int N = 5e6+100;
const double eps=1e-10;int qpow(int a,int b)
{int res=1;while(b){if(b&1) res=res*a%mod;a=a*a%mod;b>>=1;}return res;
}
int sgn(double x)
{if(fabs(x)<eps) return 0;else if(x<0) return -1;else return 1;
}
int getinv(int a){return qpow(a,mod-2LL);}
int n,a[N];
signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);//freopen("in.txt","r",stdin);cin>>n;int l=1,r=n;while(l<r){int mid=l+r>>1;int x,y;cout<<"? "<<mid<<endl;cout.flush();cout<<"? "<<mid+1<<endl;cout.flush();cin>>x>>y;if(x<y) r=mid;else l=mid+1;}cout<<"! "<<l<<endl;pause;return 0;
}

2019 ccpc 哈尔滨 E 拓扑排序

先bfs一遍确定每个点被访问到的次数，然后再按照这个次数来进行拓扑排序就可以得到1操作的那些序列被用了多少次，然后再将序列的数统计出来就可以

组队训练记录（6）：2019CCPC哈尔滨_工口发动机的博客-CSDN博客

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define lowbit(x) ((x)&(-x))
//#define int long long
#define ll long long
#define pause system("pause")
const int mod=998244353;
const int inf=1e18;
const int N = 2e6+100;
const double eps=1e-10;int qpow(int a,int b)
{int res=1;while(b){if(b&1) res=res*a%mod;a=a*a%mod;b>>=1;}return res;
}
int sgn(double x)
{if(fabs(x)<eps) return 0;else if(x<0) return -1;else return 1;
}
int getinv(int a){return qpow(a,mod-2LL);}
int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
int head[N],cnt;
struct Edge
{int next,to;
}e[N];
void addedge(int from,int to)
{e[++cnt].to=to;e[cnt].next=head[from];head[from]=cnt;
};
vector<int>g[N];
unordered_map<int,ll>mp;
int fg[N],vis[N],in[N],t,n;
ll maxx,sum,f[N];
void topo()
{queue<int>q;q.push(n);while(!q.empty()){int u=q.front();q.pop();// if(vis[u]) continue;// vis[u]=1;for(int i=head[u];i;i=e[i].next){int j=e[i].to;in[j]++;if(vis[j]) continue;q.push(j);vis[j]=1;}}q.push(n);f[n]=1;while(!q.empty()){int u=q.front();q.pop();for(int i=head[u];i;i=e[i].next){int j=e[i].to;f[j]+=f[u];in[j]--;if(in[j]<=0) q.push(j);}if(fg[u]){for(auto x:g[u]){mp[x]+=f[u];}}}
}
signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);//freopen("in.txt","r",stdin);cin>>t;while(t--){cin>>n;for(int i=1;i<=n;i++){fg[i]=vis[i]=f[i]=head[i]=in[i]=0;e[i].next=e[i].to=0;g[i].clear();}mp.clear();cnt=0;maxx=0;sum=0;for(int i=1;i<=n;i++){int op;cin>>op;if(op==1){int k;cin>>k;fg[i]=1;for(int j=1;j<=k;j++){int x;cin>>x;g[i].push_back(x);}}else{int x,y;cin>>x>>y;addedge(i,x);addedge(i,y);}}topo();for(auto x:mp){//cout<<x.second<<endl;sum+=x.second;maxx=max(maxx,x.second);}ll ans=0;if(maxx>(sum-maxx)) ans=2*(sum-maxx);else ans=sum;cout<<ans<<endl;}pause;return 0;
}

2019 ccpc 哈尔滨 I

如果h[i]>h[i-1]时答案要乘2，因为第i个数一定是当前排列的最大值或者最小值，然后有很多数变成了中间值，也就是放到i后面不会影响h[i]后面的值，这些数用一个变量num储存起来，每次乘2后，num+=(h[i]-h[i-1]-1),当h[i]==h[i-1]时就表示要放中间数了，就从num取一个

2019ccpc哈尔滨（补题） - LightAc - 博客园 (cnblogs.com).

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define lowbit(x) ((x)&(-x))
#define int long long
//#define ll long long
#define pause system("pause")
const int mod=1e9+7;
const int inf=1e18;
const int N = 2e6+100;
const double eps=1e-10;int qpow(int a,int b)
{int res=1;while(b){if(b&1) res=res*a%mod;a=a*a%mod;b>>=1;}return res;
}
int sgn(double x)
{if(fabs(x)<eps) return 0;else if(x<0) return -1;else return 1;
}
int getinv(int a){return qpow(a,mod-2LL);}
int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
int t,n,a[N];
signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);//freopen("in.txt","r",stdin);cin>>t;while(t--){int ans=1,flag=1,num=0;cin>>n;for(int i=1;i<=n;i++){cin>>a[i];if(a[1]!=0) flag=0;if(a[i]<a[i-1]||a[i]>=n) flag=0;}if(flag==0){cout<<"0\n";continue;}for(int i=2;i<=n;i++){if(a[i]>a[i-1]){ans=ans*2LL%mod;num+=(a[i]-a[i-1]-1);}else {ans=ans*num%mod;num--;}}cout<<ans<<endl;}pause;return 0;
}