Atcoder ABC144 Gluttony（贪心+二分）

Problem Statement

 

Takahashi will take part in an eating contest. Teams of $\mathrm{NN}$ members will compete in this contest, and Takahashi's team consists of $\mathrm{NN}$ players numbered $11$ through $\mathrm{NN}$ from youngest to oldest. The consumption coefficient of Member $\mathrm{ii}$ is ${\mathrm{AiAi}}^{}$.

In the contest, $\mathrm{NN}$ foods numbered $11$ through $\mathrm{NN}$ will be presented, and the difficulty of Food $\mathrm{ii}$ is ${\mathrm{FiFi}}^{}$. The details of the contest are as follows:

• A team should assign one member to each food, and should not assign the same member to multiple foods.
• It will take $\mathrm{x\times yx\times y}$ seconds for a member to finish the food, where $\mathrm{xx}$ is the consumption coefficient of the member and $\mathrm{yy}$ is the difficulty of the dish.
• The score of a team is the longest time it takes for an individual member to finish the food.

Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by $11$, as long as it does not go below $00$. However, for financial reasons, the $\mathrm{NN}$ members can do at most $\mathrm{KK}$ sets of training in total.

What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?

Constraints

 

• All values in input are integers.
• $\mathrm{1\le N\le 2\times 1051\le N\le 2\times 105}$
• $\mathrm{0\le K\le 10180\le K\le 1018}$
• $\mathrm{1\le Ai\le 106 (1\le i\le N)1\le Ai\le 106 (1\le i\le N)}$
• $\mathrm{1\le Fi\le 106 (1\le i\le N)1\le Fi\le 106 (1\le i\le N)}$

Input

 

Input is given from Standard Input in the following format:

$\mathrm{NN}$ $\mathrm{KK}$
${\mathrm{A1A1}}^{}$ ${\mathrm{A2A2}}^{}$ $......$ ${\mathrm{ANAN}}^{}$
${\mathrm{F1F1}}^{}$ ${\mathrm{F2F2}}^{}$ $......$ ${\mathrm{FNFN}}^{}$

Output

 

Print the minimum possible score of the team.

Sample Input 1

 

3 5
4 2 1
2 3 1

Sample Output 1

 

2

They can achieve the score of $22$, as follows:

• Member $11$ does $44$ sets of training and eats Food $22$ in $(4-4)\times 3=0(4-4)\times 3=0$ seconds.
• Member $22$ does $11$ set of training and eats Food $33$ in $(2-1)\times 1=1(2-1)\times 1=1$ second.
• Member $33$ does $00$ sets of training and eats Food $11$ in $(1-0)\times 2=2(1-0)\times 2=2$ seconds.

They cannot achieve a score of less than $22$, so the answer is $22$.

Sample Input 2

 

3 8
4 2 1
2 3 1

Sample Output 2

 

0

They can choose not to do exactly $\mathrm{KK}$ sets of training.

Sample Input 3

 

11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2

Sample Output 3

 

12
大意是给两个序列a1~an,f1~fn，可以任意地将ai与fj一一对应，同时有一个数k，可将a序列的某些数减小（减小的值累计不超过k，ai不能为负数）求如何操作后ai*fj的最大值最小。
最大值最小问题，很裸的二分。当时做的时候总想成是对ai*fj求和，只贪心就不太可以...二分的话由于ai*fi有单调性，所以直接二分答案，left=0，right=1e12。最关键的是check函数。首先用到贪心的思想，肯定是拿尽可能小的ai和尽可能大的fj配对，要不然大的和大的配最大值肯定会更大，所以先把a序列由小到大sort，f序列由大到小sort，check的时候检查如果当前ak*fk>mid,则减去那一部分，同时累加cnt变量（向上取整），最后看cnt是否大于k即可。

#include<iostream> 
#include<cmath>
#include<algorithm>
using namespace std;
int n;
long long int k;
long long a[200005],f[200005];
bool cmp(int a,int b)
{
    return a>b;
}
int check(long long mid)
{
    long long cnt=0;
    int i;
    for(i=1;i<=n;i++)
    {
        if(a[i]*f[i]>mid)
        {
            if((a[i]*f[i]-mid)%f[i]==0)cnt+=(a[i]*f[i]-mid)/f[i];
            else cnt+=(a[i]*f[i]-mid)/f[i]+1;
        }
    }
    return cnt<=k?1:0;
}
int main()
{
    cin>>n>>k;
    int i;
    for(i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
    }
    for(i=1;i<=n;i++)
    {
        scanf("%lld",&f[i]);
    }
    sort(a+1,a+n+1);
    sort(f+1,f+n+1,cmp);
    long long l=0,r=1e12,mid;
    while(l<r)
    {
        mid=(l+r)/2;
        if(check(mid))
        {
            r=mid;
        }
        else
        {
            l=mid+1;
        }
    }
    cout<<l;
    return 0; 
}