CodeForces

本题在比赛的时候也是想不出啊,随便贪了一下果然没用,然后也没有尝试了.
是一个很妙的贪心啊,这题是d也不是没有理由.
大概办法就是对a的序号排序(根据对应元素大小),然后分奇偶讨论,主体就是每组两个来考虑,选b中大的那个,然后处理下头就行了

/*  xzppp  */#include <iostream>#include <vector>#include <set>#include <queue>#include <map>#include <algorithm>#include <stdio.h>#include <string.h>#include <list>using namespace std;#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mp make_pairconst int MAXN = 100000;const int INF = 0x7ffffff;const int MOD = 1e9+7;typedef long long  LL;typedef unsigned long long ULL;int a[MAXN+17],b[MAXN+17],id[MAXN+17];inline bool cmp(int c,int d){    return a[c-1]>a[d-1];}int main(){    //FFF    int n;    cin>>n;    for (int i = 0; i < n; ++i) scanf("%d",a+i);    for (int i = 0; i < n; ++i) scanf("%d",b+i);    for (int i = 0; i < n; ++i) id[i]=i+1;    sort(id, id+n,cmp);    printf("%d\n",n/2+1 );    printf("%d\n",id[0] );    if(!(n&1)) printf("%d\n",id[1]);    int t = (n&1)?n/2:n/2-1,j;    for (j=n-1; t > 0; t--,j-=2)        printf("%d\n",(b[id[j]-1]>b[id[j-1]-1]?id[j]:id[j-1]));    return 0;}