拿物品

拿物品
https://ac.nowcoder.com/acm/contest/3003/F

本题实际考虑贪心
对于任意一个物品，倘若牛牛拿了，则牛牛得到ai,牛可乐失去bi，差值为ai+bi
故而每次牛牛和牛可乐都会选择 ai+bi当前最大的

#pragma warning (disable :4996)
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;

int n;
pair<ll, ll>a[maxn], b[maxn];
pair<ll, ll>Sort[maxn];
int A[maxn], B[maxn];
int cmp(pair<ll,ll> a, pair<ll,ll> b) {
    if (a.first > b.first) {
        return 1;
    }
    pair<ll, ll>c;
    c = a;
    a = b;
    b = c;
    return 0;
}
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i].first);
        a[i].second = i;
    }
    for (int j = 1; j <= n; j++) {
        scanf("%lld", &b[j].first);
        b[j].second = j;
    }
    for (int i = 1; i <= n; i++) {
        Sort[i].first = a[i].first + b[i].first;
        Sort[i].second= i;
    }
    sort(Sort + 1, Sort + n + 1, cmp);
    int ans = 0, cnt = 0;
    for (int i = 1; i <= n; i++) {
        A[ans++] = Sort[i].second;
        i++;
        if (i > n)
            break;
        B[cnt++] = Sort[i].second;
    }
    for (int i = 0; i < ans-1; i++)
        cout << A[i] << " ";
    cout <<A[ans-1]<< endl;
    for (int i = 0; i < cnt-1; i++)
        cout << B[i] << " ";
    cout <<B[cnt-1]<< endl;
}