Manthan, Codefest 16 D. Fibonacci-ish（暴力、規律）

題意:

Fibonacci−ish Sequence:=f0,f1任意,對於n≥0滿足fn+2=fn+1+fn
給定N≤103的序列,|ai|≤109,重排後,問最長的Fibonacci−ish前綴長度

分析:

由於Fibonacci數列的長度是log級的,所以直接枚舉f0,f1然後構造出這個數列,不斷更新答案就可以了
需要注意的是,特判f0=f1=0的情況,不然會退化到n3的
需要注意的是,不能重新添加map,要記錄用過的數回溯一下,不然也會退化到n3的
由於用map記錄數是否用過,時間復雜度爲O(n2log2n)

代碼:

//
//  Created by TaoSama on 2016-02-27
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include <unordered_map>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, a[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        map<int, int> mp;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            ++mp[a[i]];
        }

        int ans = mp[0];
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= n; ++j) {
                if(i == j || !a[i] && !a[j]) continue;
                vector<int> dummy;
                dummy.push_back(a[i]);
                dummy.push_back(a[j]);
                --mp[a[i]]; --mp[a[j]];
                int pre = a[j], nxt = a[i] + a[j], cnt = 2;
                //because Fibonacci sequence increases so fast, about O(log)
                while(true) {
                    if(mp[nxt]) {
                        --mp[nxt]; ++cnt;
                        dummy.push_back(nxt);
                        int tmp = nxt;
                        nxt += pre;
                        pre = tmp;
                    } else break;
                }
                ans = max(ans, cnt);
                for(int x : dummy) ++mp[x];
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}