Monster

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1383 Accepted Submission(s): 363

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<n<=10000), m (0<m<=100), indicates the number of the monsters and v11's ATK . The next n lines, each line has two integers hp (0<hp<=20), g(0<g<=1000) ,indicates the monster's HP and ATK.

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

Author

v11

Source

排序一下就解决了，注意用long long 就可以了。

1 /* *********************************************** 2 Author        :kuangbin 3 Created Time  :2013-11-17 21:24:55 4 File Name     :E:\2013ACM\比赛练习\2013-11-17\G.cpp 5 ************************************************ */ 6  7 #include 
     
       8 #include 
      
        9 #include 
       
        10 #include 
        
         11 #include 
         
          12 #include 
          
           13 #include 
           
            14 #include 
            15 #include 
             
              16 #include 
              
               17 #include 
               
                18 #include 
                
                 19 using namespace std;20 21 struct Node22 {23 int h,g;24 }node[10010];25 int n,m;26 bool cmp(Node a,Node b)27 {28 return a.h*b.g < b.h*a.g;29 }30 int main()31 {32 //freopen("in.txt","r",stdin);33 //freopen("out.txt","w",stdout);34 int T;35 int iCase = 0;36 scanf("%d",&T);37 while(T--)38 {39 iCase ++;40 scanf("%d%d",&n,&m);41 for(int i = 0;i < n;i++)42 {43 scanf("%d%d",&node[i].h,&node[i].g);44 node[i].h = (node[i].h + m - 1)/m;45 }46 sort(node,node+n,cmp);47 long long ans = 0;48 int cnt = 0;49 for(int i = 0;i < n;i++)50 {51 cnt += node[i].h;52 ans += (long long)cnt*node[i].g;53 }54 printf("Case #%d: %I64d\n",iCase,ans);55 }56 return 0;57 }

转载地址：http://ygtna.baihongyu.com/

你可能感兴趣的文章