Editorial Codeforces Round #Pi

题目：点击打开链接

567A - Lineland Mail

 分析：找与点x距离最远和最近的距离是多少？

#include<iostream>#include<stdio.h>#include<string>#include<algorithm>#include<string.h>using namespace std;int a[100005];int n,minn,maxn;int main(){    while(cin>>n){        for(int i=1;i<=n;i++)            cin>>a[i];        cout<<a[2]-a[1]<<' '<<a[n]-a[1]<<endl;        for(int i=2;i<n;i++){            maxn=max((a[n]-a[i]),(a[i]-a[1]));            minn=min((a[i+1]-a[i]),(a[i]-a[i-1]));            printf("%d %d\n",minn,maxn);        }        cout<<a[n]-a[n-1]<<' '<<a[n]-a[1]<<endl;    }    return 0;}

567B - Berland National Library

分析：求图书馆的最大容纳量，用一个集合表示现在在图书馆观众的人数，如果是+，则加入集合，如果是 -，如果在集合中没有出现过，说明以前就存在，那么maxn++，如果已经存在了，那么从集合中删除来维持现在图书馆中的数量。

#include<iostream>#include<stdio.h>#include<string>#include<algorithm>#include<string.h>#include<set>using namespace std;int a[102],n;bool in[102];int main(){    char c;    int x;    while(cin>>n){        set<int>lib;        int maxn=0;        for(int i=0;i<n;i++){            cin>>c>>x;            if(c=='+'){               lib.insert(x);               int d=lib.size();                maxn=max(maxn,d);            }            else{                if(lib.find(x)==lib.end())maxn++;                else lib.erase(x);                 int d=lib.size();                maxn=max(maxn,d);            }        }        cout<<maxn<<endl;    }    return 0;}

567C - Geometric Progression

分析：这题需要注意数据的范围( - 10⁹ ≤ a_i ≤ 10⁹），虽然a[i]不会超过int，但是a[i]*k会爆int，所以直接用LL a[2e5+5]就可以，因为这个搞了3个小时，谨记。。。

Let's solve this problem for fixed middle element of progression. This means that if we fix elementa_i then the progression must consist ofa_i / k anda_i·k elements. It could not be possible, for example, ifa_i is not divisible byk ().

For fixed middle element one could find the number of sequences by counting how manya_i / k elements are placed left from fixed element and how manya_i·k are placed right from it, and then multiplying this numbers. To do this, one could use two associative arraysA_l andA_r, where for each keyx will be stored count of occurences of x placed left (or right respectively) from current element. This could be done with map structure.

Sum of values calculated as described above will give the answer to the problem.

#include<iostream>#include<stdio.h>#include<string>#include<algorithm>#include<string.h>#include<map>using namespace std;#define LL long longLL a[200005],n,k,maxk,num;map<LL,int>lmp,rmp;void solve(){    lmp[a[1]]++;    for(int i=n;i>=2;i--)        rmp[a[i]]++;   long long ans=0;    for(int i=2;i<n;i++)    {        rmp[a[i]]--;        if(a[i]%k==0){            ans+=(long long)lmp[a[i]/k]*rmp[a[i]*k];           // cout<<lmp[a[i]/k]<<' '<<rmp[a[i]*k]<<endl;        }        lmp[a[i]]++;    }    cout<<ans<<endl;}int main(){    cin>>n>>k;    for(int i=1;i<=n;i++)        scanf("%I64d",&a[i]);    solve();    return 0;}/*3 1100001 110000 -784901888#include<iostream>#include<stdio.h>#include<string>#include<algorithm>#include<string.h>#include<map>using namespace std;#define LL long longint a[200005],n,k,maxk,num;map<LL,int>lmp,rmp;void solve(){    lmp[a[1]]++;    for(int i=n;i>=2;i--){         rmp[a[i]]++;         cout<<rmp[a[i]]<<' '<<a[i]<<endl;    }   long long ans=0;    for(int i=2;i<n;i++)    {        rmp[a[i]]--;        if(a[i]%k==0){           // cout<<(LL)a[i]*k<<endl;            ans+=(long long)lmp[a[i]/k]*rmp[(LL)a[i]*k]; //如果是int,就在这里处理一下           // cout<<lmp[a[i]/k]<<' '<<rmp[a[i]*k]<<endl;        }        lmp[a[i]]++;    }    cout<<ans<<endl;}int main(){    cin>>n>>k;    for(int i=1;i<=n;i++)        scanf("%d",&a[i]);    solve();    return 0;}