【鏈接】：CF
【題意】：對任意一個數a[i] ，可以對任意 滿足 i != j 且 a[i] > a[j] && a[i] <= a[j] +k 的 a[j] 可以被刪掉，求使最終剩下的個數最少。
【分析】：掃一遍，二分搜索合法的。
【代碼】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;

typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int maxn = 1e6+ 100;
const int maxm = 100;
const double PI = acos(-1.0);
const double eps = 1e-8;
//const int dx[] = {-1,1,0,0,1,1,-1,-1};
//const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dir[6][3]={ {0,0,1},{0,0,-1},{-1,0,0},{1,0,0},{0,1,0},{0,-1,0} };
int n,k;
int a[maxn], b[maxn];
int main()
{
    while(cin>>n>>k)
    {
        int cnt=0;
        for(int i=0;i<n;i++) cin>>a[i];/*
        for(int i=0;i<n;i++)
        {
            cout<<i<<"   ";
        }
        cout<<endl;*/
        sort(a,a+n,greater<int>());/*
        for(int i=0;i<n;i++)
        {
            cout<<a[i]<<' ';
        }
        cout<<endl;*/
        for(int i=0;i<n;i++)
        {
            int p = lower_bound(a,a+n,a[i]+k,greater<int>())-a;   //第一個 <= (a[i]+k) 的位置
            //cout<<"a[i]+k="<<a[i]+k<<" p="<<p<<" a[p]="<<a[p]<<" a[i]="<<a[i];
            if(a[p]>a[i])
            {
                cnt++;
                //cout<<" YES";
            }
            //cout<<endl;
        }
        cout<<n-cnt<<endl;
    }
}
/*
7 1
101 53 42 102 101 55 54
42 53 54 55 101 101 102
outputCopy
3
42 55 102
inputCopy
6 5
20 15 10 15 20 25
25
outputCopy
1
inputCopy
7 1000000
1 1 1 1 1 1 1
outputCopy
7


7 1
101 53 42 102 101 55 54
101 53 42 102 101 55 54
102 101 101 55 54 53 42
a[i]+k=103 p=0 a[p]=102 a[i]=102
a[i]+k=102 p=0 a[p]=102 a[i]=101 YES
a[i]+k=102 p=0 a[p]=102 a[i]=101 YES
a[i]+k=56 p=3 a[p]=55 a[i]=55
a[i]+k=55 p=3 a[p]=55 a[i]=54 YES
a[i]+k=54 p=4 a[p]=54 a[i]=53 YES
a[i]+k=43 p=6 a[p]=42 a[i]=42
*/
 

[模擬]

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;

typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int maxn = 1e6+ 100;
const int maxm = 100;
const double PI = acos(-1.0);
const double eps = 1e-8;
//const int dx[] = {-1,1,0,0,1,1,-1,-1};
//const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dir[6][3]={ {0,0,1},{0,0,-1},{-1,0,0},{1,0,0},{0,1,0},{0,-1,0} };
int n,k,j;
int a[maxn], b[maxn];
int main()
{
    while(cin>>n>>k)
    {
        int cnt, j=0;
        for(int i=0;i<n;i++) cin>>a[i];

        sort(a,a+n);
        cnt=n;
        for(int i=0;i<n;i++)
        {
           while(a[j]<a[i])
           {
               if(a[j]+k>=a[i]) cnt--;
               j++;
           }
        }
        cout<<cnt<<endl;
    }
}
 

CF 990B B. Micro-World【數組操作/貪心/STL/二分搜索】