二分 Codeforces633D Fibonacci-ish

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题意：找出最长的斐波那契数列，f0和f1可以是出现的数字里的任何数

思路：枚举f0和f1，根据斐波那契的收敛性，我们能顺着去判断出最长有多长，注意要特判掉f0和f1都等于0的情况，不然复杂度会退化到O(n^3)

刚开始是用map，然后T了，其实用数组搞一搞就行了，这里用了一种很傻逼的写法来清空数组，和线段树的懒惰标记很类似

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck(x) cout<<"["<<x<<"]"#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;const int MX = 1e3 + 5;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;int cnt[MX], sign[MX];int A[MX], B[MX], C[MX], sz = 0;int ID(int x) {    int l = 1, r = sz, m;    while(l <= r) {        m = (l + r) >> 1;        if(C[m] == x) return m;        if(C[m] < x) l = m + 1;        else r = m - 1;    }    return 0;}bool check(int p, int tim) {    if(sign[p] != tim) {        sign[p] = tim;        cnt[p] = 0;    }    return true;}int main() {    //FIN;    int n; scanf("%d", &n);    for(int i = 1; i <= n; i++) {        scanf("%d", &A[i]);    }    sort(A + 1, A + 1 + n);    for(int l = 1, r; l <= n; l = r + 1) {        for(r = l; r + 1 <= n && A[r + 1] == A[r]; r++);        C[++sz] = A[l]; B[sz] = r - l + 1;    }    int ans = max(B[ID(0)], 2);    for(int i = 1; i <= n; i++) {        for(int j = 1; j <= n; j++) {            if(i == j || (!A[i] && !A[j])) continue;            int f0 = A[i], f1 = A[j], f2 = f0 + f1, id = ID(f2), now = 2;            int id1 = ID(A[i]), id2 = ID(A[j]), tim = i * n + j;            check(id1, tim); check(id2, tim); cnt[id1]++; cnt[id2]++;            while(id && check(id, tim) && cnt[id] + 1 <= B[id]) {                sign[id] = i * n + j;                cnt[id]++; now++;                f0 = f1; f1 = f2;                f2 = f0 + f1;                id = ID(f2);            }            ans = max(ans, now);        }    }    printf("%d\n", ans);    return 0;}

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