题目链接（hdu复赛）
2018年湘潭邀请赛，在湘大举行，当时一起打的过去的，因为没出市就方便点。第一次参加现场赛，首先反省一下自己，比赛前算法先不说(没有好好搞过)，就连普通的题，数学性质的题都很少刷，什么都不会，然后英语不说四六级的东西，很多专业英语词汇都没有好好了解过，主要的原因还是刷题刷少了，英文题刷少了。理所当然，成功拿下一铁，湘潭赛打铁告终。
比赛开始，这次是三个人没人一份纸质题目，拿到题目，看最后一题，k.2018发现可以做，(事实证明确实是一道水题)，我就在做这题，他们看了a题，好像是k题一顿操作后提交，错了，一直到比赛结束都没做出来，后来回去看别人的题解，发现自己情况没有分析全面，其实是完全可以解出来的，真的做的太少太少了。a题最后也是没过，还有一个f题，sort，我最初的想法是用stl里的那个sort排序，只要对sort的compare函数做处理应该可以完成排序，比赛之前我看过，但是不熟悉，zxm她也看了我就交给她了，最后好像因为爆long double的问题也没做出来。
哎，菜还是菜，很多算法都不懂，数据结构也没学好，很有一段时间我都特别头疼算法，不想学，费劲，觉得自己不适合学计算机，更不适合ACM。有时候又想，不适合好像总是loser的借口！总是在后悔和偷懒的矛盾中！
A. Easy 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 published many papers. Given a0,a1,a2,…,an which means Bobo has published ai papers with
itations exactly i, find the h-index of Bobo.
Input
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains an integer n. The second line contains (n+1) integers a0,a1,…,an.
Output
For each test case, print an integer which denotes the result.
Constraint
• 1≤ n ≤2·105
• 0≤ ai ≤109
• The sum of n does not exceed 250,000.
Sample Input
1
1 2
2
1 2 3
3
0 0 0 0
Sample Output
1
2
0
题意：给定被引用次数为0~n的论文分别有几张，找到最大的h，满足被引用次数大于等于h的论文至少有h张
思路：在区间[0,n]内二分答案；或直接从n~0遍历找到第一个满足条件的h
后AC代码
#include "bits/stdc++.h"
using namespace std;
int main(){
int a[200005];
int n;
int i;
while(cin>>n){
for(i=0;i<=n;i++)
cin>>a[i];
int sum=a[n];
for(i=n;i>=0;){
if(sum>=i){
cout<<i<<endl;
break;
}
else sum+=a[--i];
}
if(i<0)
cout<<"0"<<endl;
}
return 0;
}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.
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
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
#include<cstdio>
int main(){
int n,a;
while(scanf("%d%d",&n,&a)!=EOF){
printf("%d\n",a+(n-a)/2);
}
return 0;
}
Sample Input
2
1 1 1
1 1 2
2
1 1 2
1 1 1
3
1 3 1
2 2 1
3 1 1
Sample Output
2 1
1 2
1 2 3
题意：给定n个元组(a1,b1,c1),(a2,b2,c2),…,(an,bn,cn)，将其按(ai+bi)/(ai+bi+ci)的值从小到大排序，输出排序后的n个元组的原序号；
思路：编写sort里的cmp函数(形参为元组结构体元素，设为Tuple x,Tuple y)，若直接算出(x.a+x.b)(y.a+y.b+y.c)和(y.a+y.b)(x.a+x.b+x.c)再比较大小，这两个结果会爆unsigned long long；
可以把因式乘积展开成多项式的和，约去两式中相同的项，得到x.a*y.c+x.b*y.c和y.a*x.c+y.b*x.c，因此只需计算它俩再比较即可，结果不会爆unsigned long long
后AC代码
#include "bits/stdc++.h"
using namespace std;
struct node{
long double a,b,c;
int numb;
}ss[1005];
bool cmp(const node &a,const node &b){
long double suma,sumb;
//suma=a.a*b.c+a.b*b.c;
//sumb=b.a*a.c+b.b*a.c;
suma=(a.a+a.b)/(a.a+a.b+a.c);
sumb=(b.a+b.b)/(b.a+b.b+b.c);
if(suma!=sumb)return suma<sumb;
return a.numb<b.numb;
}
int main(){
int n;
while(cin>>n){
for(int i=0;i<n;i++){
cin>>ss[i].a>>ss[i].b>>ss[i].c;
ss[i].numb=i+1;
}
stable_sort(ss,ss+n,cmp);
int i;
for(i=0;i<n-1;i++)
cout<<ss[i].numb<<" ";
cout<<ss[i].numb<<endl;
}
return 0;
}