Monster

http://acm.hdu.edu.cn/showproblem.php?pid=3979

Problem Description
One day, v11 encounters a group of monsters in a foreast. In order to defend the homeland, V11 picks up his weapon and fights!
All the monsters attack v11 at the same time. Every enemy has its HP, and attack value ATK. In this problem, v11 has his ATK and infinite HP. The damage (also means reduction for HP) is exactly the ATK the attacker has. For example, if v11's ATK is 13 and the monster's HP is 27, then after v11's attack, the monster's HP become 27 - 13 = 14 and vice versa.
v11 and the monsters attack each other at the same time and they could only attack one time per second. When the monster's HP is less or equal to 0 , we think this monster was killed, and obviously it would not attack any more. For example, v11's ATK is 10 and a monster's HP is 5, v11 attacks and then the monster is killed! However, a monster whose HP is 15 will be killed after v11 attack for two times. v11 will never stop until all the monsters are killed ! He wants to minimum the HP reduction for the fight! Please note that if in some second, some monster will soon be killed , the monster's attack will works too.
 
Input
The first line is one integer T indicates the number of the test cases. (T <=100)
Then for each case, The first line have two integers n (0 
Output
Output one line.
First output “Case #idx: ”, here idx is the case number count from 1. Then output the minimum HP reduction for v11 if he arrange his attack order optimal .
 
Sample Input

   
   
   
   

    
    
    
    2 
3 1 
1 10 
1 20 
1 40 
1 10 
7 3
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    Case #1: 110 
Case #2: 3
   
   
   
   

욕심은 우선 모든 몬스터를 죽이는 데 필요한 시간 t를 계산한 다음에 몬스터의 피해와 살해에 필요한 비율에 따라 순서를 정한다. 물론 비율이 높은 몬스터를 우선적으로 죽인다.
롱롱을 사용해야 합니다.

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
struct node
{
    LL hp;
    LL g;
    LL t;
}monster[10005];
bool cmp(const struct node &a,const struct node &b)
{
    return (double)a.g/a.t>(double)b.g/b.t;
}
int main()
{
    int T,n,m;
    scanf("%d",&T);
    for(int tt=1;tt<=T;++tt)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;++i)
        {
            scanf("%I64d%I64d",&monster[i].hp,&monster[i].g);
            monster[i].t=(monster[i].hp/m+((monster[i].hp%m)?1:0));
        }
        sort(monster,monster+n,cmp);
        LL reduction=0,t=0;
        for(int i=0;i<n;++i)
        {
            t+=monster[i].t;
            reduction+=(t*monster[i].g);
        }
        printf("Case #%d: %I64d
",tt,reduction);
    }
    return 0;
}

출처:http://blog.csdn.net/acm_ted/article/details/7773137