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You are going out for a walk, when you suddenly encounter N monsters. Each monster has a parameter called health, and the health of the i-th monster is h_i at the moment of encounter. A monster will vanish immediately when its health drops to 0 or below.

Fortunately, you are a skilled magician, capable of causing explosions that damage monsters. In one explosion, you can damage monsters as follows:

• Select an alive monster, and cause an explosion centered at that monster. The health of the monster at the center of the explosion will decrease by A, and the health of each of the other monsters will decrease by B. Here, A and B are predetermined parameters, and A>Bholds.

At least how many explosions do you need to cause in order to vanish all the monsters?

Constraints

• All input values are integers.
• 1≤N≤10⁵
• 1≤B<A≤10⁹
• 1≤h_i≤10⁹

Input

Input is given from Standard Input in the following format:

N A B
h1
h2
:
hN

Output

Print the minimum number of explosions that needs to be caused in order to vanish all the monsters.

Sample Input 1

4 5 3
8
7
4
2

Sample Output 1

2

You can vanish all the monsters in two explosion, as follows:

• First, cause an explosion centered at the monster with 8 health. The healths of the four monsters become 3, 4, 1 and ?1, respectively, and the last monster vanishes.
• Second, cause an explosion centered at the monster with 4 health remaining. The healths of the three remaining monsters become 0, ?1 and ?2, respectively, and all the monsters are now vanished.

Sample Input 2

2 10 4
20
20

Sample Output 2

4

You need to cause two explosions centered at each monster, for a total of four.

Sample Input 3

5 2 1
900000000
900000000
1000000000
1000000000
1000000000

Sample Output 3

800000000

二分啊啊啊  这么大的数 不二分怎么写

 1 #include <iostream>
 2 using namespace std;
 3 #include<string.h>
 4 #include<set>
 5 #include<stdio.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<map>
 9 #include<algorithm>
10 #include<cstdio>
11 #include<cmath>
12 #include<cstring>
13 #include <cstdio>
14 #include <cstdlib>
15 #include<stack>
16 #include<vector>
17 long long  a[51000000];
18 long long  n,a1,b1;
19 long  panduan(long long  zhi)
20 {
21     long long  t=b1*zhi;
22     long long  add=0;
23     for(int i=1;i<=n;i++)
24     {
25         if(a[i]-t>0)
26         {
27             if((a[i]-t)%(a1-b1)==0)
28                 add+=(a[i]-t)/(a1-b1);
29             else
30                 add+=(a[i]-t)/(a1-b1)+1;
31         }
32     }
33     //cout<<zhi<<"_"<<add<<endl;
34     return add<=zhi;
35 }
36 int  main()
37 {
38     cin>>n>>a1>>b1;
39     for(int i=1;i<=n;i++)
40         scanf("%d",&a[i]);
41     long long  kaishi=0,jieshu=1e9;
42     long long  mid;
43     //cout<<panduan(3)<<"_"<<endl;
44     int t;
45     while(kaishi<jieshu-1)
46     {
47         //cout<<kaishi<<"_"<<jieshu<<endl;
48         mid=(jieshu+kaishi)>>1;
49         if(panduan(mid))
50         {
51             jieshu=mid;
52             t=mid;
53         }
54         else
55             kaishi=mid;
56     }
57     cout<<jieshu<<endl;
58     return 0;
59 }

View Code

AtCoder - 2580 Widespread

标签：inpu   ring   orm   least   val   imu   standard   cti   can   

原文地址：http://www.cnblogs.com/dulute/p/7966703.html