Problem 62 min

$des $
There are $n $ items, the first $i $ items have two attributes $k _i, b_i$, which represents it at the moment $x $ 's value for $k _i
\times x + b_i$.
Currently at the moment $0$, you can choose not to exceed $m $ items, making the existence of an integer moment $t $, $t >= 0$
, the total value of all items you select is greater than or equal to $S $.
Given $S $, the minimum value for $t $.

$sol $
Select any of the collections, and the resulting gain and all can be represented as a function of a single form. Only care about these
The maximum value of a function can be found that the maximum value must be decreased first (when T is very large, $k _i \times t > b_i$, it may be monotonically increasing or monotonically decreasing).
So just check to see if the 0-hour meeting is eligible, and if not, two points.
Note that the check is only needed to find the largest m, so you can do O (n), the specific way is to quickly
In the process of recursion only one side. Directly with the STL nth_element ().
Time complexity $O (nlog10^9) $

$code $

#include <bits/stdc++.h>#defineRep (I, J, K) for (int i = j; I <= K; i++)using namespacestd;intRead () {Charc = GetChar ();intx =0; intsig =1;  while(C <'0'|| C >'9') {if(c = ='-') sig =-1; c =GetChar ();}  while(c >='0'&& C <='9') x = x *Ten+ C-'0', C =GetChar (); returnX *Sig;}Const intN = 1e6 +Ten; typedefLong LongLL;intN, M; LL S;intK[n], b[n]; LL Val[n];BOOLCheckintx) {Rep (I,1, n) val[i] = 1LL * K[i] * x +B[i]; Nth_element (Val+1, Val + M, Val + n +1, greater<ll>()); LL sum=0; Rep (i,1, m)if(Val[i] >0&& (sum + = Val[i]) >= S)return true; return false;}intMain () {n= Read (), M = Read (); scanf"%lld", &S); Rep (i,1, n) k[i] = Read (), b[i] =Read (); if(Check (0) {puts ("0");return 0;} intL =1, R =1e9, Ans;  while(L <=R) {intMid = (L + R)/2; if(Check (mid)) Ans = Mid, R =mid; ElseL = mid +1; } printf ("%d\n", Ans); return 0;}

Problem 62 min