二分答案加双指针扫描 gym101194D

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 300005;
ll a[maxn],b[maxn];
int t,T,n,k;
bool judge(int x)    //第一层总是优先选取较小的为最优策略
{
    for(int i = 0 ;i< x;i++)
        a[i] = b[i];
    int p = x;
    for(int i =x ;i< x*k;i++)
    {
        while(b[p]<a[i-x]*2 && p<n) p++;
        if(p==n)
            return 0;
        a[i]=b[p];
        p++;
    }
    return 1;


}
int bitsection(int l,int r)
{
    while(l<r)
    {
        int mid = (l+r+1)>>1;
        if(judge(mid))
            l=mid;
        else r=mid-1;


    }
    return l;
}


int main()
{
    cin>>T;
    for(t=1;t<=T;t++)
    {
        cin>>n>>k;
        for(int i = 0; i < n; i++)
            cin>>b[i];
        sort(b,b+n);
        ll ans =bitsection(0,n/k);          //满足二分性质
        printf("Case #%d: %lld\n",t,ans);
    }
    return 0;
}