Codefroces D2. Magic Powder - 2(二分)
http://codeforces.com/problemset/problem/670/D2
http://codeforces.com/problemset/problem/670/D1
1 second
256 megabytes
standard input
standard output
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
二分模板
#include<iostream>
using namespace std;
int binary_search(int a[],int l,int r,int k)
{
int mid,ans;
while(l<=r)
{
mid=(l+r)>>1;
if(a[mid]<=k){
l=mid+1;
ans=mid;
}
else r=mid-1;
}
return ans;//也可ans+n;
}
int main()
{
int a[11]={1,2,4,7,7,7,7,8,10,11};
cout<<binary_search(a,0,9,4)<<endl;
return 0;
}
题解
#include<iostream>
#include<cstdio>
using namespace std;
long long a[],b[];
int main()
{
long long n,k,i;
scanf("%lld%lld",&n,&k);
for(i=;i<n;i++)
{
scanf("%lld",&a[i]);
}
for(i=;i<n;i++)
{
scanf("%lld",&b[i]);
}
long long l=,r=;
while(l<=r)
{
long long mid=(l+r)>>;
long long sum=;
for(i=;i<n;i++)
{
if(mid*a[i]>b[i]) sum+=(mid*a[i]-b[i]);
if(sum>k) break;
}
if(sum<=k) l=mid+;
else r=mid-;
}
printf("%lld",l-);
return ;
}
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