思路：
二分最大化平均值。
迭代次数太多会超时。
实现：
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 const int MAXN = 100005; 5 const int INF = 0x3f3f3f3f; 6 int v[MAXN], w[MAXN], ans[MAXN], n, k; 7 struct node 8 { 9 int id; 10 double x; 11 }; 12 node buf[MAXN]; 13 bool cmp(const node & a, const node & b) 14 { 15 return a.x < b.x; 16 } 17 bool check(double x) 18 { 19 for (int i = 0; i < n; i++) { buf[i].x = v[i] - x * w[i]; buf[i].id = i; } 20 sort(buf, buf + n, cmp); 21 double sum = 0; 22 for (int i = n - 1; i > n - 1 - k; i--) sum += buf[i].x; 23 if (sum >= 0) 24 { 25 for (int i = n - 1; i > n - 1 - k; i--) ans[n - 1 - i] = buf[i].id; return true; 26 } 27 return false; 28 } 29 int main() 30 { 31 scanf("%d %d", &n, &k); 32 for (int i = 0; i < n; i++) scanf("%d %d", &v[i], &w[i]); 33 double l = 0, r = INF; 34 for (int i = 0; i < 50; i++) 35 { 36 double m = (l + r) / 2; 37 if (check(m)) l = m; 38 else r = m; 39 } 40 for (int i = 0; i < k; i++) printf("%d ", ans[i] + 1); 41 return 0; 42 }