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

CODEFORCES-670D2 Magic 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 and K (1≤n≤100 000, 1≤k≤109)-the number of ingredients and the N Umber of grams of the magic powder.

The second line contains the sequence a1, A2, ..., an (1≤ai≤109), where the i-th number was equal to the number of gram s 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 would be a able to bake using the ingredients that she have and the mag IC Powder.

Sample Input Input

1 1000000000
1
1000000000

Output

2000000000

Input

1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000< C5/>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

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 100010
using 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;
			}
			else
				r=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 100010 using 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;
		} else r=mid-1; } priNTF ("%lld\n", ans);
} return 0; }