Codeforces-670d2 Magic Powder-2 (dichotomy & Simulation)

Source: Internet
Author: User

codeforces-670d2Magic Powder-2
Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %i64d &%i64u

Submit Status

Description

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

Input

The first line contains the positive integers n andK (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 o f Thei-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 Thei-th ingredient, which Apollinaria has.

Output

Print the maximum number of cookies, which Apollinaria would be a able to bake using the ingredients that she have and the mag IC Powder.

Sample Input

Input
1 100000000011000000000
Output
2000000000
Input
10 11000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 10000000001 1 1 1 1 1 1 1) 1 1
Output
0
Input
3 12 1 411 3 16
Output
4
Input
4 34 3 5 611 12 14 20
Output
3

Sample Output

Hint

Source

Codeforces Round #350 (Div. 2)
Test instructions: The previous question, is the number opened big, must use two points hait:inf relatively small, therefore must use 3000000000LL on the line
#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> #define LL Long Long#define INF 0x3f3f3f3f#define N 100010using namespace Std;ll a[n],b[n];int n;ll k;bool judge (ll X) {ll s=k;for (int i=1; i<=n;i++) {if (X*a[i]>=b[i]) s-= (X*a[i]-b[i]), if (s<0) return false;} return true;} int main () {int i;while (scanf ("%d%lld", &n,&k)!=eof) {for (i=1;i<=n;i++) scanf ("%lld", &a[i]); for (i=1;i <=n;i++) scanf ("%lld", &b[i]), if (n==1) {printf ("%lld\n", (K+b[1])/a[1]); continue;} ll L=0,r=3000000000ll,mid,ans;while (l<=r) {mid= (l+r) >>1ll;if (judge (mid)) {L=mid+1;ans=mid;} Elser=mid-1;} printf ("%lld\n", ans);} return 0;}

This is just beginning to write, more trouble.
#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> #define LL Long Long#define INF 0x3f3f3f3f#define N 100010using namespace std;ll a[n],b[n];struct zz{ll a;ll b;ll c;ll D;} P[n];bool CMP (ZZ a,zz b) {return A.A<B.A;} int N;ll K;bool judge (ll X) {ll s=0;for (int i=1;i<=n;i++) {if (p[i].a<x) s+=p[i].c+ (x-p[i].a-1) *p[i].d;if (s>k) return false;} return true;} int main () {int i;while (scanf ("%d%lld", &n,&k)!=eof) {for (i=1;i<=n;i++) {scanf ("%lld", &a[i]);p [i].d=a [i];} for (i=1;i<=n;i++) scanf ("%lld", &b[i]), if (n==1) {printf ("%lld\n", (K+b[1])/a[1]); continue;} for (i=1;i<=n;i++) {p[i].a=b[i]/a[i];p [i].b=b[i]%a[i];p [i].c=a[i]-p[i].b;} Sort (p+1,p+n+1,cmp); ll L=0,r=3000000000ll,mid,ans;while (l<=r) {mid= (l+r) >>1ll;if (judge (mid)) {l=mid+1; Ans=mid;} Elser=mid-1;} printf ("%lld\n", ans);} return 0;}



Codeforces-670d2 Magic Powder-2 (dichotomy & Simulation)

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.