poj 1840 Eqs

　　简单hash。。。。很想知道那些200ms以内的是怎么搞的。。。

View Code

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 const int mod = 4000037;
 8 const int HASH = 0x20decade;
 9 const int inf = 0x7fffffff;
10 
11 int hash[mod];
12 int used[mod] = {0};
13 int tri[101];
14 
15 void pre() {
16     for (int i = -50; i <= 50; i++) {
17         tri[i + 50] = i * i * i;
18     }
19 }
20 
21 void insert(int _x) {
22     int p = ((_x ^ HASH) & inf) % mod;
23 
24     while (hash[p] != _x && used[p]) p = (p + 1) % mod;
25     hash[p] = _x;
26     used[p]++;
27 }
28 
29 int find(int _x) {
30     int p = ((_x ^ HASH) & inf) % mod;
31 
32     while (hash[p] != _x && used[p]) p = (p + 1) % mod;
33 
34     return used[p];
35 }
36 
37 int main() {
38     int a[5];
39     bool nz = false;
40 
41     pre();
42     for (int i = 0; i < 5; i++) {
43         scanf("%d", &a[i]);
44         if (a[i]) nz = true;
45     }
46     if (!nz) {
47         puts("10000000000");
48 
49         return 0;
50     }
51     for (int i = 0; i <= 100; i++) {
52         if (i == 50) continue;
53         int t1 = tri[i] * a[0];
54 
55         for (int j = 0; j <= 100; j++) {
56             if (j == 50) continue;
57             int t2 = tri[j] * a[1];
58 
59             for (int k = 0; k <= 100; k++) {
60                 if (k == 50) continue;
61                 insert(t1 + t2 + tri[k] * a[2]);
62             }
63         }
64     }
65     __int64 cnt = 0;
66 
67     for (int i = 0; i <= 100; i++) {
68          if (i == 50) continue;
69          int t = tri[i] * a[3];
70 
71          for (int j = 0; j <= 100; j++) {
72              if (j == 50) continue;
73              cnt += find(- (t + tri[j] * a[4]));
74          }
75     }
76 
77     printf("%I64d\n", cnt++);
78 
79     return 0;
80 }

——written by Lyon

posted @ 2012-10-01 00:12 LyonLys 阅读(180) 评论(0) 编辑收藏举报

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