题解 CF940A 【Points on the line】

此题之水，水于上青天

$(1<=n<=100,0<=d<=1001<=n<=100,0<=d<=100 )$

$n$ 只有100！

思路：

先~~sort~~排序，再双重循环枚举，求出范围 $i$ ~ $j$ ，其余剩下的（不在这个范围内的），比如 $a[i-1]$ ，就是删去的。然后打擂台：

m=min(m,n-j+i-1);//算出i~j里面有多少个数字

最后输出m即可。

AC CODE：

#include<iostream>
#include<algorithm>//sort头文件
using namespace std;
int n,m=100000000;//m一定要开大一点！
int d;
int a[30000];
int main(){cin>>n>>d;for(int i=1;i<=n;i++){cin>>a[i];}sort(a+1,a+n+1);//排序for(int i=1;i<=n;i++){for(int j=i;j<=n;j++){//可能存在i=j(最大值和最小值相同的情况)if(a[j]-a[i]<=d){m=min(m,n-j+i-1);//打擂台}}}cout<<m;return 0;//华丽的结束
}

~~码风新奇勿喷~~

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题解 CF940A 【Points on the line】

此题之水，水于上青天

$(1<=n<=100,0<=d<=1001<=n<=100,0<=d<=100 )$

思路：

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