Codeforces - Mike and distribution

题目链接：Codeforces - Mike and distribution

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此题可以随机化，比较玄学。
AC代码：

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e5+10;
std::mt19937 rnd(time(0));
int n,k,flag,a[N],b[N],p[N];	long long sa,sb;
signed main(){
	cin>>n;	 k=n/2+1;
	for(int i=1;i<=n;i++)	scanf("%d",&a[i]),sa+=a[i],p[i]=i;
	for(int i=1;i<=n;i++)	scanf("%d",&b[i]),sb+=b[i];
	while(!flag){
		shuffle(p+1,p+1+n,rnd);	 long long s1=0,s2=0;
		for(int i=1;i<=k;i++){
			s1+=a[p[i]],s2+=b[p[i]];
			if(s1>sa/2&&s2>sb/2){flag=1;	break;}
		}
	}
	cout<<k<<endl;
	for(int i=1;i<=k;i++)	printf("%d ",p[i]);
	return 0;
}

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正解：
我们按照a从大到小排序。
然后看相邻的两个b，取大的一个，那么必然合法。

AC代码：

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e5+10;
int n,k;
struct node{int a,b,id;}t[N];
int cmp(node s1,node s2){return s1.a>s2.a;}
signed main(){
	cin>>n;	k=n/2+1;
	for(int i=1;i<=n;i++)	scanf("%d",&t[i].a),t[i].id=i;
	for(int i=1;i<=n;i++)	scanf("%d",&t[i].b);
	sort(t+1,t+1+n,cmp);
	printf("%d\n%d",k,t[1].id);
	for(int i=2;i<=n;i+=2)
		if(t[i+1].b>t[i].b)	printf(" %d",t[i+1].id);
		else	printf(" %d",t[i].id);
	return 0;
}