Codeforces 567A 567B 567C 567D

All cities of Lineland are located on the Ox coordinate axis. Thus, each city is associated with its position x_i — a coordinate on the Ox axis. No two cities are located at a single point.

Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).

Strange but true, the cost of sending the letter is exactly equal to the distance between the sender‘s city and the recipient‘s city.

For each city calculate two values ??min_i and max_i, where min_i is the minimum cost of sending a letter from the i-th city to some other city, and max_i is the the maximum cost of sending a letter from the i-th city to some other city

#include<iostream>
#include<cstdio>
#include<cstring>
#include<math.h>
#include<algorithm>
using namespace std;
#define N 500005
typedef long long ll;
#include <queue>
#include <set>
#include <map>
#include <stack>
set<int>q;
set<int>::iterator it,itt;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        char s[101][2];
        int x;
        int ans=0;
        for(int i=0; i<n; i++)
        {
            cin>>s[i]>>x;
            if(s[i][0]=='-')
            {
                if(q.count(x))
                    q.erase(x);
                else
                    ans++;
            }
            else
            {
                q.insert(x);
                int ans1=q.size();
                ans=max(ans,ans1);
            }
        }
         cout<<ans<<endl;
    }
    return 0;
}

#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <map>
#include <cstdio>
#include <set>
#include <vector>
using namespace std;
__int64 n,k;
__int64 a[200005];
__int64 b[200005]; //记录i位置的数对应 有多少个 a[i]/k 存在
map <int ,__int64 > c;
int main()
{
    __int64 i;
    scanf("%I64d%I64d",&n,&k);
    for (i=1; i<=n; i++)
    {
        scanf("%I64d",&a[i]);
        if (a[i]%k==0)
            b[i]=c[a[i]/k]; 		//记录当前位置 拥有多少个 a[i]/k (等比的前一项)
        c[a[i]]++;
        //每次把a[i]个数加一
    }
    // for(i=1; i<=n; i++)
    // cout<<b[i]<<endl;
    c.clear();
    __int64 ans=0;
    for (i=1; i<=n; i++)
    {
        if (a[i]%k==0)
            ans+=c[a[i]/k];   //a[i]为第三项，a[i]/k为第二项，c[a[i]/k]存的是（每一个（第二项对应的第一项）的个数）之和
        c[a[i]]+=b[i]; 	//c[a[i]] 存的是 数a[i] 对应的 a[i]/k 的个数
    }
    printf("%I64d\n",ans);
    return 0;
}

Alice and Bob love playing one-dimensional battle ships. They play on the field in the form of a line consisting of n square cells (that is, on a 1?×?n table).

At the beginning of the game Alice puts k ships on the field without telling their positions to Bob. Each ship looks as a 1?×?a rectangle (that is, it occupies a sequence of a consecutive squares of the field). The ships cannot intersect and even touch each other.

After that Bob makes a sequence of "shots". He names cells of the field and Alice either says that the cell is empty ("miss"), or that the cell belongs to some ship ("hit").

But here‘s the problem! Alice like to cheat. May be that is why she responds to each Bob‘s move with a "miss".

Help Bob catch Alice cheating — find Bob‘s first move, such that after it you can be sure that Alice cheated.

#include<cstdlib>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define maxn 1100000
using namespace std;
int bb[maxn],a[maxn],n,m,k,aa;
bool check(int x)
{
    int size=0,ans=0;
    for(int i=1; i<=n; i++)
    {
        size++;
        if(bb[i]<=x&&bb[i])size=0;
        if(size>=aa)
        {
            size=-1;
            ans++;
        }
    }
    if(ans>=k)
    return 1;
    else
     return 0;
}
int main()
{
    scanf("%d%d%d%d",&n,&k,&aa,&m);
    for(int i=1; i<=m; i++)
    {
        scanf("%d",&a[i]);
        bb[a[i]]=i;
    }
    int l=1,r=m;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(check(mid))l=mid+1;
        else r=mid-1;
    }
    if(l<=m)printf("%d\n",l);
    else puts("-1");
    return 0;
}

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