hdu 4379 The More The Better ( 2012 Multi-University Training Contest 8)

  http://acm.hdu.edu.cn/showproblem.php?pid=4379

The More The Better

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 800    Accepted Submission(s): 208

Problem Description

Given an sequence of numbers {X ₁, X ₂, ... , X _n}, where X _k = (A * k + B) % mod. Your task is to find the maximum sub sequence {Y ₁, Y ₂, ... , Y _m} where every pair of (Y _i, Y _j) satisfies Y _i + Y _j <= L (1 ≤ i < j ≤ m), and every Y_i <= L (1 ≤ i ≤ m ).
Now given n, L, A, B and mod, your task is to figure out the maximum m described above.

 

Input

Multiple test cases, process to the end of input. Every test case has a single line. A line of 5 integers: n, L, A, B and mod. (1 ≤ n ≤ 2*10 ⁷, 1 ≤ L ≤ 2*10 ⁹, 1 ≤ A, B, mod ≤ 10 ⁹)

 

Output

For each case, output m in one line.

 

Sample Input

   
     

      
    1 8 2 3 6 
   
     
5 8 2 3 6

 

Sample Output

   
     

      
    1 4
   
     

 

Source

2012 Multi-University Training Contest 8

 

 

 

简单题，首先想到所有小于 L/2 的，统统可以放进来，最后，按照题意，还可能可以放一个大于 L/2 的数进来，当小于 L/2 的数里面的最大值加上这个大于 L/2 的数的和小于 L 时，答案加一。最后要注意所有数都小于 L/2 的处理。O(n) 算法可过此题。

 

好想说这是好蛋疼的一道题，错了不知道多少次 ，发现 错在了  min = 1<<62  和 min = 1   min<<= 62  的值竟然不一样（min 为 long long）tmd。。。。。

 

View Code

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<queue>
 7 #include< set>
 8 #include<map>
 9  #define Min(a,b)  a>b?b:a
10  #define Max(a,b)  a>b?a:b
11  #define CL(a,num)  memset(a,num,sizeof(a));
12  #define inf  1<<62
13  #define maxn    1000010
14  #define eps  1e-6
15  #define ll  __int64
16 
17 
18  using  namespace std;
19 
20 
21  int main()
22 {
23 
24     ll a,b;
25 
26     ll mod,n;
27     ll l;
28       int  k;
29       // freopen("data.txt","r",stdin);
30       while(scanf( " %I64d%I64d%I64d%I64d%I64d ",&n,&l,&a,&b,&mod)!=EOF)
31     {
32          ll mx =  0;
33          int cnt =  0;
34         ll  mi = inf ; 
35           int mid = l/ 2;
36          for( k =  1 ; k <= n;++k)
37         {
38              ll tmp = ( a*k + b)%mod ;
39              if(tmp <= mid)
40             {
41                 cnt ++;
42                  if(tmp > mx) mx = tmp;
43 
44             }
45              else
46             {
47 
48                  if(mi > tmp) mi = tmp ;
49 
50 
51             }
52 
53         }
54 
55 
56          if(mi + mx <= l)cnt++;
57         printf( " %d\n ",cnt);
58     }
59      return  0;
60 
61 }