Save the Nature
题意：
在第n * a张卖掉的电影票能够提供x%的收益
在第n * b张卖掉的电影票能够提供y%的收益
问收益为k时，最少需要卖多少张电影票
思路：
二分一下答案就好了
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<iostream> 6 #include<vector> 7 #include<queue> 8 #define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) 9 #define mem(a,x) memset(a,x,sizeof(a)) 10 #define lson rt<<1,l,mid 11 #define rson rt<<1|1,mid + 1,r 12 #define P pair<ll,ll> 13 #define ull unsigned long long 14 using namespace std; 15 typedef long long ll; 16 const int maxn = 2e5 + 10; 17 const int inf = 0x3f3f3f3f; 18 ll q, n, k; 19 ll val[maxn]; 20 ll tmp[maxn]; 21 ll x, y, a, b; 22 priority_queue<P, vector<P>, less<P>>que, tque; 23 int gcd(int a,int b){while (b ^= a ^= b ^= a %= b); return a;} 24 int lcm(int a, int b) { return a * b / gcd(a, b); } 25 26 bool check(int mid) 27 { 28 tque = que; 29 ll pos = n, sum = 0; 30 ll cnt = 0; 31 while (!tque.empty() && sum < k && pos > 0) 32 { 33 P cur = tque.top(); 34 tque.pop(); 35 if (cur.second > mid) continue; 36 sum += val[pos--] / 100 * cur.first; 37 ++cnt; 38 if (cnt >= mid) break; 39 } 40 return sum >= k; 41 } 42 43 int main() 44 { 45 scanf("%lld", &q); 46 while (q--) 47 { 48 while (!que.empty()) que.pop(); 49 bool flag = false; 50 scanf("%lld", &n); 51 for (int i = 1; i <= n; ++i) 52 { 53 scanf("%lld", &val[i]); 54 } 55 sort(val + 1, val + 1 + n); 56 scanf("%lld %lld", &x, &a); 57 scanf("%lld %lld", &y, &b); 58 scanf("%lld", &k); 59 mem(tmp, 0); 60 for (int i = 1; i * a <= n; ++i) 61 tmp[i * a] += x; 62 for (int i = 1; i * b <= n; ++i) 63 tmp[i * b] += y; 64 for (int i = 1; i <= n; ++i) 65 if (tmp[i] > 0) 66 que.push(P(tmp[i], i)); 67 int l = 1, r = n + 1; 68 while (r > l) 69 { 70 int mid = (l + r) >> 1; 71 if (check(mid)) 72 { 73 flag = true; 74 r = mid; 75 } 76 else 77 { 78 l = mid + 1; 79 } 80 } 81 if (flag) printf("%d ", r); 82 else printf("-1 "); 83 } 84 85 return 0; 86 }