标签：
http://codeforces.com/problemset/problem/670/D2
The term of this problem is the same as the previous one, the only exception — increased restrictions.
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.
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
1 1000000000
1
1000000000
2000000000
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
0
3 1
2 1 4
11 3 16
4
4 3
4 3 5 6
11 12 14 20
3
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
using namespace std;
#define N 110000
#define met(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
const long long Max = 2000000000;
typedef long long LL;
LL a[N], b[N];
LL n, k;
LL Judge(LL mid)
{
LL i, K1=k, K2=k;
for(i=1; i<=n; i++)
{
if(b[i]<a[i]*mid)
{
K1 -= (a[i]*mid - b[i]);
if(K1<0)
return -1;
}
}
for(i=1; i<=n; i++)
{
if(b[i]<a[i]*(mid+1))
{
K2 -= (a[i]*(mid+1) - b[i]);
if(K2<0)
return 0;
}
}
return 1;
}
int main()
{
while(scanf("%I64d%I64d", &n, &k)!=EOF)
{
LL i;
LL mid, L=0, R=Max, ans;
met(a, 0);
met(b, 0);
for(i=1; i<=n; i++)
scanf("%I64d", &a[i]);
for(i=1; i<=n; i++)
scanf("%I64d", &b[i]);
while(L<R)
{
mid = (L+R)/2;
ans = Judge(mid);
if(ans==0)
L = R = mid;
if(ans>0)
L = mid + 1;
if(ans<0)
R = mid - 1;
}
printf("%I64d\n", L);
}
return 0;
}
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原文地址：http://www.cnblogs.com/YY56/p/5466361.html
