poj 2785 4 Values whose Sum is 0

　　简单hash，不过不能用set，而且数组不能开太大。

　　才几十行的代码，当时个人赛居然不会... - -

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 1 #include <cstdlib>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 const int maxn = 4001;
 8 const int mod = 15999997;
 9 int hash[maxn * maxn], cnt[maxn * maxn];
10 int rec[maxn][4];
11 
12 void insert(int n){
13     int p = (n << 2) % mod;
14 
15     if (p < 0) p += mod;
16     while (cnt[p] && hash[p] != n) p++;
17     hash[p] = n;
18     cnt[p]++;
19 }
20 
21 int count(int n){
22     int p = (n << 2) % mod;
23 
24     if (p < 0) p += mod;
25     while (cnt[p] && hash[p] != n) p++;
26     return cnt[p];
27 }
28 
29 int deal(int n){
30     memset(cnt, 0, sizeof(cnt));
31     for (int i = 0; i < n; i++){
32         for (int j = 0; j < n; j++){
33             insert(rec[i][0] + rec[j][1]);
34         }
35     }
36     int ret = 0;
37 
38     for (int i = 0; i < n; i++){
39         for (int j = 0; j < n; j++){
40             ret += count(- rec[i][2] - rec[j][3]);
41         }
42     }
43 
44     return ret;
45 }
46 
47 int main(){
48     int n;
49 
50     while (~scanf("%d", &n)){
51         for (int i = 0; i < n; i++){
52             for (int j = 0; j < 4; j++){
53                 scanf("%d", &rec[i][j]);
54             }
55         }
56         printf("%d\n", deal(n));
57     }
58 
59     return 0;
60 }

——written by Lyon

posted @ 2012-09-02 00:38 LyonLys 阅读(130) 评论(0) 收藏举报

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