https://codeforces.com/contest/670/problem/D2

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.

Examples

input

1 1000000000 1 1000000000

output

2000000000

input

10 1 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1 1 1 1 1 1 1 1 1 1

output

0

input

3 1 2 1 4 11 3 16

output

4

input

4 3 4 3 5 6 11 12 14 20

output

3

題意：同上一題，就是數開大了，得用二分 Hait: INF比較小，所以得用3000000000ll就行了

#include <iostream> #include<stdio.h> #include<string.h> using namespace std; typedef long long ll ; ll need[100008]; ll have[100008]; ll tmpp[100008]; ll n,k; ll Slove(ll mid) { ll sum=k;for(ll i=1;i<=n;i++){tmpp[i]=have[i]-mid*need[i];if(tmpp[i]<0)sum+=tmpp[i];if(sum<0)return 0;}return 1; } int main() {while(~scanf("%I64d%I64d",&n,&k)){for(ll i=1;i<=n;i++){scanf("%I64d",&need[i]);}for(ll i=1;i<=n;i++){scanf("%I64d",&have[i]);}ll ans=0;ll mid;ll l=0;ll r=3000000000ll;while(r>=l){mid=(l+r)/2;if(Slove(mid)==1){ans=mid;l=mid+1;}else{r=mid-1;}}printf("%I64d\n",ans);}//cout << "Hello world!" << endl;return 0; }

?

總結

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