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：
1 //#include<bits/stdc++.h> 2 #include <set> 3 #include <map> 4 #include <stack> 5 #include <cmath> 6 #include <queue> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstring> 11 #include <iostream> 12 #include <algorithm> 13 14 #define ll long long 15 #define pll pair<ll,ll> 16 #define pii pair<int,int> 17 #define bug printf("********* ") 18 #define FIN freopen("input.txt","r",stdin); 19 #define FON freopen("output.txt","w+",stdout); 20 #define IO ios::sync_with_stdio(false),cin.tie(0) 21 #define ls root<<1 22 #define rs root<<1|1 23 #define pb push_back 24 25 using namespace std; 26 const int inf = 2e9 + 7; 27 const ll Inf = 1e18 + 7; 28 const int maxn = 2e5 + 5; 29 const int mod = 1e9 + 7; 30 31 bool cmp(const ll& a, const ll& b) 32 { 33 return a > b; 34 } 35 36 ll p[maxn], n, a[maxn], k, tmp[maxn]; 37 38 bool check(ll mid) 39 { 40 for (int i = 1; i <= mid; ++i) tmp[i] = a[i]; 41 sort(tmp + 1, tmp + 1 + mid, cmp); 42 ll res = 0; 43 for (int i = 1; i <= mid; ++i) 44 { 45 res += p[i] / 100 * tmp[i]; 46 } 47 return res >= k; 48 } 49 50 int main() 51 { 52 int T; 53 scanf("%d", &T); 54 while (T--) 55 { 56 scanf("%lld", &n); 57 for (int i = 1; i <= n; ++i) 58 { 59 scanf("%lld", &p[i]); 60 a[i] = 0; 61 } 62 ll x1, y1, x2, y2; 63 scanf("%lld %lld %lld %lld", &x1, &y1, &x2, &y2); 64 scanf("%lld", &k); 65 for (int i = 1; i <= n; ++i) 66 { 67 if (i % y1 == 0) a[i] += x1; 68 if (i % y2 == 0) a[i] += x2; 69 } 70 sort(p + 1, p + 1 + n, cmp); 71 ll ans = -1; 72 ll l = 1, r = n; 73 while (l <= r) 74 { 75 ll mid = l + r >> 1; 76 if (check(mid)) 77 r = mid - 1, ans = mid; 78 else 79 l = mid + 1; 80 } 81 printf("%lld ", ans); 82 } 83 }