链接：https://ac.nowcoder.com/acm/contest/1107/B
来源：2019牛客国庆集训派对day2
- 题目描述
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.- 输入描述
The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers n and a.
1 ≤ n ≤ 109
0 ≤ a ≤ n
The number of test cases does not exceed 104.- 输出描述
For each test case, print an integer which denotes the maximum h-index.- 输入
3 0
3 1
1000000000 1000000000- 输出
1
2
1000000000- 备注
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：Bobo 印刷一些纸张，他有 n 个小时，他可以选择任意的时间段( x小时 )去印刷纸张，当前纸张的贡献是 a × x，如果当前的纸张之前有印刷好的纸张，那么当前纸张的贡献还要加上前面印刷好的纸张的数量。问最大的 h，h表示至少有 h 张纸的贡献至少为 h。
思路：我们假设每张纸都有 1 小时的时间来印刷，直接二分 h，mid >= n - mid + a ，一共印刷了 n 张纸，其中满足条件的有 mid 张，n - mid 是前面生产好的纸张加上 a 是这些纸的最小的贡献。前面的式子如果成立说明当前的贡献(n - mid + a)满足题意，并且可以更大，否则说明纸张数量过多，减小即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Max_n = 1e6 + 10;
int a, n;
bool check(int mid) {
return mid >= n - mid + a;
}
int main() {
while(~scanf("%d%d", &n, &a)) {
int l = 1, r = n, ans = 0;
while(l <= r) {
int mid = (l + r) / 2;
if(check(mid)) {
ans = n - mid + a;
r = mid - 1;
} else l = mid + 1;
}
printf("%d\n", ans);
}
return 0;
}
/**
* Copyright(c)
* All rights reserved.
* Author : Max_n
* Date : 2019-10-02-13.16.09
* Problem : Higher h-index
*/
