Save the Nature CF-1241C（二分、贪心）

题意：

有$n$件商品，每天可以卖一件，每件商品的售价是$p[i]$元，在$a$的倍数天可以获得$x$%的利润，在$b$的倍数天可以获得$y$%的利润，利润可叠加。

如果需要获得$k$的利润，问最少需要卖多少天。

思路：

二分答案。对于售价高的商品分配给前$mid$天中利润高的时间。

代码：

//#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define pb push_back

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;

bool cmp(const ll& a, const ll& b)
{
    return a > b;
}

ll p[maxn], n, a[maxn], k, tmp[maxn];

bool check(ll mid)
{
    for (int i = 1; i <= mid; ++i) tmp[i] = a[i];
    sort(tmp + 1, tmp + 1 + mid, cmp);
    ll res = 0;
    for (int i = 1; i <= mid; ++i)
    {
        res += p[i] / 100 * tmp[i];
    }
    return res >= k;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%lld", &n);
        for (int i = 1; i <= n; ++i)
        {
            scanf("%lld", &p[i]);
            a[i] = 0;
        }
        ll x1, y1, x2, y2;
        scanf("%lld %lld %lld %lld", &x1, &y1, &x2, &y2);
        scanf("%lld", &k);
        for (int i = 1; i <= n; ++i)
        {
            if (i % y1 == 0)    a[i] += x1;
            if (i % y2 == 0)    a[i] += x2;
        }
        sort(p + 1, p + 1 + n, cmp);
        ll ans = -1;
        ll l = 1, r = n;
        while (l <= r)
        {
            ll mid = l + r >> 1;
            if (check(mid))
                r = mid - 1, ans = mid;
            else
                l = mid + 1;
        }
        printf("%lld
", ans);
    }
}