hihoCoder 1239 Fibonacci

#1239 : Fibonacci

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Given a sequence {an}, how many non-empty sub-sequence of it is a prefix of fibonacci sequence.

A sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

The fibonacci sequence is defined as below:

F1 = 1, F2 = 1

Fn = Fn-1 + Fn-2, n>=3

输入

One line with an integer n.

Second line with n integers, indicating the sequence {an}.

For 30% of the data, n<=10.

For 60% of the data, n<=1000.

For 100% of the data, n<=1000000, 0<=ai<=100000.

输出

One line with an integer, indicating the answer modulo 1,000,000,007.

样例提示

The 7 sub-sequences are:

{a2}

{a3}

{a2, a3}

{a2, a3, a4}

{a2, a3, a5}

{a2, a3, a4, a6}

{a2, a3, a5, a6}

样例输入

62 1 1 2 2 3

样例输出

7

解题思路：先打表求fib数列前25项(恰好小于10^6)，然后DP

dp[i][j]表示到第i位为止，fib数列第j项的表达方法有多少种

dp[i][1]=dp[i-1][1]

dp[i][j]=dp[i-1][j]+dp[i-1][j-1]//dp[i-1][j]表示用当前项替代i-1之前的第j项，dp[i-1][j-1]表示i-1之前的第j-1项加上当前项

一定要取模，否则溢出会WA

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <vector>#include <bitset>#include <cmath>#include <utility>#define Maxn 100005#define Maxm 1000005#define MOD 1000000007#define lowbit(x) x&(-x)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define PI acos(-1.0)#define make_pair MP#define LL long long #define Inf (1LL<<62)#define inf 0x3f3f3f3f#define re freopen("in.txt","r",stdin)#define wr freopen("out.txt","w",stdout)using namespace std;int fib[30];LL doc[Maxm][26];map<int,int> _fib;void init(){fib[1]=fib[2]=1;_fib[1]=1;for(int i=3;fib[i-1]+fib[i-2]<=100000;i++)fib[i]=fib[i-1]+fib[i-2],_fib[fib[i]]=i;}int main(){init();int n,num,ans;//re;while(~scanf("%d",&n)){ans=0;memset(doc,0,sizeof(doc));scanf("%d",&num);if(num==1){ans++;doc[0][1]=1;}for(int i=1;i<n;i++){scanf("%d",&num);for(int j=1;j<=25;j++)doc[i][j]=doc[i-1][j];if(_fib[num]){if(num==1){if(doc[i][1])//fib数列第二项{ans+=doc[i-1][1];ans++;ans%=MOD;doc[i][1]=(doc[i-1][1]+1)%MOD;doc[i][2]=(doc[i-1][1]+doc[i-1][2])%MOD;}else//fib数列第一项{ans++;doc[i][1]=(doc[i-1][1]+1)%MOD;}}else{ans+=doc[i-1][_fib[num]-1];//fib数列第三项之后ans%=MOD;if(doc[i][2])doc[i][_fib[num]]=(doc[i-1][_fib[num]]+doc[i-1][_fib[num]-1])%MOD;}}}printf("%d\n",ans);}return 0;}

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