ASC7 Problem C. Fibonacci Subsequence

题目大意

给你一个长为$m(m\le 3000)$的数列，求最长Fibonacci子序列。

简要题解

DP即可，因为头两个随便选，所以考虑设$f[i][j]$表示以$a_i,a_j$开始的Fibonacci子序列最长长度为多少，只需要从$f[j][k],a_k=a_i+a_j$转移，关键在于找到$k$，map和hash_map都TLE了，最后手写了一个HashTable，Accept。

有一个细节就是转移的顺序，先从后往前枚举第一维$i$，再从$i-1$到$0$枚举第二维$j$，就可以避免每次清空HashTable。

输出答案没必要保存DP信息（不然会MLE），直接由Fibonacci前两项就可以全部推出来了，我个智障。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 namespace my_header {
  4 #define pb push_back
  5 #define mp make_pair
  6 #define pir pair<int, int>
  7 #define vec vector<int>
  8 #define pc putchar
  9 #define clr(t) memset(t, 0, sizeof t)
 10 #define pse(t, v) memset(t, v, sizeof t)
 11 #define bl puts("")
 12 #define wn(x) wr(x), bl
 13 #define ws(x) wr(x), pc(' ')
 14     const int INF = 0x3f3f3f3f;
 15     typedef long long LL;
 16     typedef double DB;
 17     inline char gchar() {
 18         char ret = getchar();
 19         for(; (ret == '\n' || ret == '\r' || ret == ' ') && ret != EOF; ret = getchar());
 20         return ret; }
 21     template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) {
 22         for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar());
 23         if (c == '-') { flg = -1; c = getchar(); }
 24         for(ret = 0; '0' <= c && c <= '9'; c = getchar())
 25             ret = ret * 10 + c - '0';
 26         ret = ret * flg; }
 27     inline int fr() { int t; fr(t); return t; }
 28     template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); }
 29     template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); }
 30     template<class T> inline char wr(T a, int b = 10, bool p = 1) {
 31         return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 
 32             (wr(a/b, b, 0), pc('0' + a % b)));
 33     }
 34     template<class T> inline void wt(T a) { wn(a); }
 35     template<class T> inline void wt(T a, T b) { ws(a), wn(b); }
 36     template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); }
 37     template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); }
 38     template<class T> inline T gcd(T a, T b) {
 39         return b == 0 ? a : gcd(b, a % b); }
 40     template<class T> inline T fpw(T b, T i, T _m, T r = 1) {
 41         for(; i; i >>= 1, b = b * b % _m)
 42             if(i & 1) r = r * b % _m;
 43         return r; }
 44 };
 45 using namespace my_header;
 46 
 47 const int MAXN = 3000 + 3;
 48 int a[MAXN], n;
 49 short f[MAXN][MAXN];
 50 
 51 struct HashTable {
 52     static const int MOD = 100019;
 53     static const int HASHSIZE = 3333;
 54     int h[MOD], nxt[HASHSIZE], key[HASHSIZE], val[HASHSIZE], cnt;
 55     HashTable() {
 56         memset(h, -1, sizeof h);
 57         cnt = 0;
 58     }
 59     int getHash(int p) {
 60         return (p % MOD + MOD) % MOD;
 61     }
 62     void insert(int k, int v) {
 63         int t = getHash(k);
 64         val[cnt] = v;
 65         key[cnt] = k;
 66         nxt[cnt] = h[t];
 67         h[t] = cnt++;
 68     }
 69     int find(int k) {
 70         int t = getHash(k);
 71         for (int i = h[t]; i != -1; i = nxt[i]) {
 72             if (key[i] == k)
 73                 return val[i];
 74         }
 75         return 0;
 76     }
 77 } pos;
 78 
 79 int main() {
 80 #ifdef lol
 81     freopen("C.in", "r", stdin);
 82     freopen("C.out", "w", stdout);
 83 #else
 84     freopen("fibsubseq.in", "r", stdin);
 85     freopen("fibsubseq.out", "w", stdout);
 86 #endif
 87     fr(n);
 88     for (int i = 1; i <= n; ++i)
 89         fr(a[i]);
 90     for (int i = n; 0 < i; --i) {
 91         for (int j = i - 1; 0 < j; --j) {
 92             int t = pos.find(a[i] + a[j]);
 93             if (f[i][t] + 1 > f[j][i])
 94                 f[j][i] = f[i][t] + 1;
 95         }
 96         pos.insert(a[i], i);
 97     }
 98     int ans = 0, ri, rj;
 99     for (int i = 1; i <= n; ++i)
100         for (int j = i + 1; j <= n; ++j)
101             if (f[i][j] >= ans) {
102                 ans = f[i][j];
103                 ri = a[i];
104                 rj = a[j];
105             }
106     wt(ans + 1);
107     if (ans == 0) {
108         wt(a[1]);
109         return 0;
110     }
111     for (int i = 0; i <= ans; ++i) {
112         ws(ri);
113         int t = ri + rj;
114         ri = rj;
115         rj = t;
116     }
117     puts("");
118 
119     return 0;
120 }

 

posted @ 2017-02-16 10:11  ichneumon  阅读(284)  评论(0)    收藏  举报