Physical Examination

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

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

 

Sample Output

1419

 

 

水题，贪心

a1*b2 < a2*b1 的情况选a

 

#include <iostream>
#include <algorithm>
using namespace std;

#define Mod (365*24*60*60)
struct S
{
    long long a,b;
};

bool cmp(S x, S y)
{
    return x.a*y.b < x.b*y.a;
}

S s[100002];

int main()
{
    int n;
    
    while (cin >> n, n)
    {
        for (int i = 0; i < n; ++i)
        {
            cin >> s[i].a >> s[i].b;
        }
        
        sort(s, s+n, cmp);
        
        long long ans = 0;
        for (int i = 0; i < n; ++i)
        {
            if (ans != 0)
            {
                ans += (s[i].b*ans)%Mod;
                
                ans %= Mod;
            }
            
            ans += s[i].a;
            
            ans %= Mod;
        }
        
        cout << ans << endl;    
        
    }
}