題目:

The term of this problem is the same as the previous one, the only exception — increased restrictions.

Input

The first line contains two positive integers?n?and?k?(1?≤?n?≤?100?000,?1?≤?k?≤?109) — the number of ingredients and the number of grams of the magic powder.

The second line contains the sequence?a1,?a2,?...,?an?(1?≤?ai?≤?109), where the?i-th number is equal to the number of grams of the?i-th ingredient, needed to bake one cookie.

The third line contains the sequence?b1,?b2,?...,?bn?(1?≤?bi?≤?109), where the?i-th number is equal to the number of grams of the?i-th ingredient, which Apollinaria has.

Output

Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.

解答：

首先很很明顯這是一個對結果進行二分的題目，

注意要用long long int，同時在判斷目標答案是否可行的時候用減法不要用加法，否則可能會溢出

#define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <map> #include <vector> #include <algorithm> #define ll long long int using namespace std;struct vege {ll id;ll per;ll total;ll alre;vege() {} };ll cost(vector<vege> &ingr, int tar,ll k) {ll res = 0;ll size = ingr.size();for (int i = 0; i < size; ++i) {if (k < 0)return false;k = k - (tar *ingr[i].per - ingr[i].total <= 0 ? 0 : tar *ingr[i].per - ingr[i].total);}if (k < 0)return false;return true; } static bool comp(vege A, vege B) {return A.alre < B.alre; } int main() {ll n, k;cin >> n >> k;vector<vege> ingr(n);ll per;for (int i = 0; i < n; ++i) {cin >> per;ingr[i].id = i;ingr[i].per = per;}for (int i = 0; i < n; ++i) {cin >> per;ingr[i].total = per;ingr[i].alre = per / ingr[i].per;}//sort(ingr.begin(), ingr.end(), comp);int l = 0;int r = ~(1 << 31);int res = 0;while (l <= r) {int mid = l + (r - l) / 2;bool co = cost(ingr, mid, k);if (co) {l = mid + 1;res = mid;}else {r = mid - 1;}}cout << res << endl;//system("pause"); }

總結

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