hdu 4442

Physical Examination

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7429    Accepted Submission(s): 2232

Problem Description

WANGPENG is a freshman. He is requested to have a physical examination when entering the university.
Now WANGPENG arrives at the hospital. Er….. There are so many students, and the number is increasing!
There are many examination subjects to do, and there is a queue for every subject. The queues are getting longer as time goes by. Choosing the queue to stand is always a problem. Please help WANGPENG to determine an exam sequence, so that he can finish all the physical examination subjects as early as possible.

 

Input

There are several test cases. Each test case starts with a positive integer n in a line, meaning the number of subjects(queues).
Then n lines follow. The i-th line has a pair of integers (ai, bi) to describe the i-th queue:
1. If WANGPENG follows this queue at time 0, WANGPENG has to wait for ai seconds to finish this subject.
2. As the queue is getting longer, the waiting time will increase bi seconds every second while WANGPENG is not in the queue.
The input ends with n = 0.
For all test cases, 0<n≤100000, 0≤a_i,b_i<2³¹.

 

Output

For each test case, output one line with an integer: the earliest time (counted by seconds) that WANGPENG can finish all exam subjects. Since WANGPENG is always confused by years, just print the seconds mod 365×24×60×60.

 

Sample Input

51 22 33 44 55 60

 

Sample Output

1419
Hint
In the Sample Input, WANGPENG just follow the given order. He spends 1 second in the first queue, 5 seconds in the 2th queue, 27 seconds in the 3th queue, 169 seconds in the 4th queue, and 1217 seconds in the 5th queue. So the total time is 1419s. WANGPENG has computed all possible orders in his 120-core-parallel head, and decided that this is the optimal choice.
 

 

Source

2012 Asia JinHua Regional Contest 

题意：

某人去医院检查身体，有n个部门

刚开始的时候 time == 0  ，然后开始计时

每个部门就诊需要 a  秒 ，并且需要等待 time * b秒

求最短的时间

解题
假设有两个队伍a和b，如先去排a，则总体等待时间为：a1+a1*b2+a2
如先去排b,则总体的等待时间为a1+a2*b1+a2
我们假设a在前优，则  a1*b2  <  a2*b1 
由此推广到多个队伍，按照ax*by<ay*bx的标准排序后所得的顺序处理即为答案

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn=100010;const int mod=365*24*60*60;struct node{    long long a,b;}que[maxn];int cmp(const node x,const node y){    return (x.a)*(y.b) < (x.b)*(y.a);}int main(){    long long n;    //freopen("in.txt","r",stdin);    while(scanf("%lld",&n)!=EOF)    {        if(n==0)            break;        for(int i=0;i<n;i++)            scanf("%lld%lld",&que[i].a,&que[i].b);        sort(que,que+n,cmp);        long long ans=0;        for(int i=0;i<n;i++){            ans=ans+que[i].a+ans*que[i].b;            ans%=mod;        }        printf("%lld\n",ans);    }    return 0;}