Physical Examination

//假设有两组队，最短时间时排序满足a1*b2<a2*b1;所有数组按这个从小到大排即为最快，同时要注意sum的类型

A - Physical Examination

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4442

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<2 ³¹.

 

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.          

 

代码如下//

#include <cstdio>#include <cstring>#include <algorithm>#define Max 100000+5using namespace std;#define mod 31536000;int n;struct que{    int len;    int t;}f[Max];bool cmp(que a, que b){ return (long long)a.len*b.t < (long long)a.t*b.len;}int main(){    int i, j;    while(scanf("%d",&n) && n){        for(i = 0; i < n; i++)            scanf("%d%d",&f[i].len,&f[i].t);        sort(f,f+n,cmp);        long long sum = 0;        for(i = 0; i < n; i++){            if(f[i].len == 0) continue;            sum += (long long)(f[i].len+(long long)sum*f[i].t);            sum %= mod;        }        printf("%lld\n",sum);    }}