标签：example you isp info bool ons size ash input
You have array a1, a2, ..., an. Segment [l, r] (1 ≤ l ≤ r ≤ n) is good if ai = ai - 1 + ai - 2, for all i (l + 2 ≤ i ≤ r).
Let‘s define len([l, r]) = r - l + 1, len([l, r]) is the length of the segment [l, r]. Segment [l1, r1], is longer than segment [l2, r2], if len([l1, r1]) > len([l2, r2]).
Your task is to find a good segment of the maximum length in array a. Note that a segment of length 1 or 2 is always good.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of elements in the array. The second line contains integers: a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
Print the length of the longest good segment in array a.
Examples
10
1 2 3 5 8 13 21 34 55 89
10
5
1 1 1 1 1
2
sol：找最长的满足斐波那契数列性质的数列，容易发现只要55个数字就会数字大小就会爆int，但是如果你直接暴力的话100000个0你就T飞了
所以把一串0缩成一个点，在暴力
但是有一堆地方要特判，我跪的很惨（我太菜菜菜菜菜菜菜菜菜菜了）
#include <bits/stdc++.h> using namespace std; typedef int ll; inline ll read() { ll s=0; bool f=0; char ch=‘ ‘; while(!isdigit(ch)) { f|=(ch==‘-‘); ch=getchar(); } while(isdigit(ch)) { s=(s<<3)+(s<<1)+(ch^48); ch=getchar(); } return (f)?(-s):(s); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar(‘-‘); x=-x; } if(x<10) { putchar(x+‘0‘); return; } write(x/10); putchar((x%10)+‘0‘); return; } #define W(x) write(x),putchar(‘ ‘) #define Wl(x) write(x),putchar(‘\n‘) const int N=100005; int n; int a[N],A[N],Len[N]; int main() { int i,j,ans=1; R(n); if(n<=2) {Wl(n); return 0;} for(i=1;i<=n;i++) { R(a[i]); } *A=0; for(i=1;i<=n;i++) { if(a[i]>0) { A[++*A]=a[i]; Len[*A]=1; } else { A[++*A]=0; for(;i<=n&&a[i]==0;i++) Len[*A]++; i--; } } for(i=1;i<=n;i++) ans=max(ans,Len[i]); if(*A==1) ans=n; if(*A==2) { if(A[1]==0) ans=max(Len[1],Len[2]+1); else ans=Len[2]; } for(i=1;i<=(*A)-2;i++) { int tmp; if(A[i]==0) { tmp=Len[i+1]+1; } else if(A[i+1]==0) { if(Len[i+1]==1) tmp=Len[i+1]+1; else { tmp=1+Len[i+2]; for(j=i+3;j<=*A;j++) { if(A[j]==A[j-1]+A[j-2]) tmp+=Len[j]; else break; } ans=max(ans,tmp); continue; } } else tmp=Len[i]+Len[i+1]; for(j=i+2;j<=*A;j++) { if(A[j]==A[j-1]+A[j-2]) tmp+=Len[j]; else break; } ans=max(ans,tmp); } Wl(ans); return 0; } /* input 10 1 2 3 5 8 13 21 34 55 89 output 10 input 5 1 1 1 1 1 output 2 input 10 1 1 0 0 0 0 0 0 0 1 output 7 */
标签：example you isp info bool ons size ash input
原文地址：https://www.cnblogs.com/gaojunonly1/p/10617439.html
