Higher h-index+CCPC2018-湖南全国邀请赛

当（n+a）/2为偶数时，此数最小
Higher h-index

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 152    Accepted Submission(s): 91

Problem Description

The h-index of an author is the largest h where he has at least h papers with citations not less than h.

Bobo has no papers and he is going to publish some subsequently.
If he works on a paper for x hours, the paper will get (a⋅x) citations, where a is a known constant.
It's clear that x should be a positive integer.
There is also a trick -- one can cite his own papers published earlier.

Given Bobo has n working hours, find the maximum h-index of him.

 

Input

The input consists of several test cases and is terminated by end-of-file.

Each test case contains two integers n and a.

 

Output

For each test case, print an integer which denotes the maximum h-index.

## Constraint

* 1≤n≤109
* 0≤a≤n
* The number of test cases does not exceed 104.

 

Sample Input

3 0 3 1 1000000000 1000000000

 

Sample Output

1 2 1000000000

Hint

For the first sample, Bobo can work $3$ papers for $1$ hour each. With the trick mentioned, he will get papers with citations $2, 1, 0$. Thus, his $h$-index is $1$. For the second sample, Bobo can work $2$ papers for $1$ and $2$ hours respectively. He will get papers with citations $1 + 1, 2 + 0$. Thus, his $h$-index is $2$.

 

Source

CCPC2018-湖南全国邀请赛-重现赛（感谢湘潭大学）

 

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liuyiding

#define happy

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

typedef pair<int,int> pi;
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typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;


#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)

#define all(a) (a).begin(),(a).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define f first
#define s second

ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int N=1e4+10;

int a,b;

int main(){
#ifdef happy
    freopen("in.txt","r",stdin);
#endif
    int n;
    while(~scanf("%d%d",&a,&b)){
        printf("%d\n",(a+b)/2);
    }
}