B.Higher h-index

1 B. Higher h-index

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.

1.1 Input

The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers n and a.

1.2 Output

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

1.3 Constraint

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

1.4 Sample Input

3 0
3 1
1000000000 1000000000

1.5 Sample Output

1
2
1000000000

1.6 Note

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.

题意：给定 n 个小时，可以用其中 x(1<=x<=n) 个小时写一篇论文，那么这篇论文的"既定"引用数将会是x*a(a 为给定正整数）；此外，已经写好的论文将会被其后写成的论文所引用，也就是说，这篇论文的总引用数将会是"既定"引用数 + 其后论文篇数；问在所有的写论文方案中（例如一种方案就是用 n 个小时写 n 篇论文，每篇论文各花 1 小时（可以得到这 n 篇论文的引用数）)，h 最大为多少 (h 的含义同上题）（每一种方案都对应着一个 h，求这些 h 中的最大者）
思路：最优方案（即对应 h 值最大的方案）是平摊 n 小时写成 n 篇论文（证明未知）；此时 n 篇论文的引用数为 a,a+1,a+2,…,a+n-1，引用数为 a+i 时，引用数大于等于它的论文有 n-i 篇，令 a+i=n-i 得 i=(n-a)/2, 所以 h=a+(n-a)/2;

后 AC 代码

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#include<cstdio>

int main(){
   int n,a;
   while(scanf("%d%d",&n,&a)!=EOF){
    printf("%d\n",a+(n-a)/2);
   }
   return 0;
}