二分 Codeforces633D Fibonacci-ish

传送门：点击打开链接

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;
}