2015
07-16

# Physical Examination

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.

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≤ai,bi<231.

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≤ai,bi<231.

5
1 2
2 3
3 4
4 5
5 6
0

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.


/*

贪心(2012金华现场赛A题)。
假设已经排了t时间了，现在有两个队列，a1、a2、b1、b2。

化简后，只剩下了b2*a1和a2*b1，那么按照这个，对每两个进行

2012-10-30
*/

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
struct A
{
__int64 a,b;
}E[100011];
int cmp(const void *a,const void *b)
{
struct A *c,*d;
c=(struct A *)a;
d=(struct A *)b;
__int64 t1,t2;
t1=(c->a)*(d->b);
t2=(d->a)*(c->b);
if(t1>t2)   return 1;
else        return -1;
}
int main()
{
int n;
int i,l;
int temp;
__int64 ans;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)    scanf("%I64d%I64d",&E[i].a,&E[i].b);
qsort(E,n,sizeof(E[0]),cmp);

ans=0;
temp=365*24*60*60;
for(i=0;i<n;i++)    {ans+=E[i].a+ans*E[i].b;ans%=temp;}
printf("%I64d\n",ans);
}
return 0;
}