Codeforces Round #350 (Div. 2)_D2 - Magic Powder - 2
D2. Magic Powder - 2
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputThe 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
题解:二分答案的经典应用
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
![](https://images.cnblogs.com/OutliningIndicators/ExpandedBlockStart.gif)
1 #include<cstdio> 2 #define LL long long 3 #define FFC(i,a,b) for(int i=a;i<=b;i++) 4 LL a[100010],b[100010],n,k,maxn=2*1e9+2; 5 void fre(){freopen("c:\\acm\\input.txt","r",stdin);} 6 LL search(LL l,LL r){ 7 LL mid,sum; 8 while(l<=r){ 9 mid=(l+r)>>1,sum=0; 10 FFC(i,1,n){ 11 sum+=a[i]*mid-b[i]>0?a[i]*mid-b[i]:0; 12 if(sum>k)break;//防爆LL 13 } 14 if(sum==k)return mid; 15 else if(sum<k)l=mid+1; 16 else r=mid-1; 17 } 18 return r; 19 } 20 int main(){ 21 //fre(); 22 while(~scanf("%I64d%I64d",&n,&k)){ 23 FFC(i,1,n)scanf("%I64d",&a[i]); 24 FFC(i,1,n)scanf("%I64d",&b[i]); 25 printf("%I64d\n",search(1,maxn)); 26 } 27 return 0; 28 }