Codeforces 126D Fibonacci Sums 求n由任意的Sum(fib)的方法数 dp

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
#define ll __int64
#define N 90
ll n ,m;
ll f
, dp
[2]; //dp[i][0]表示不拆分下标为i的数 dp[i][1]表示拆分下标为i的数
vector<int>G;
int main(){  
	ll i, j, u, v, T; cin>>T;
	f[1] = 1; f[2] = 2;
	for(i=3;i<N;i++)f[i] = f[i-1] + f[i-2];
	while(T--) {
		G.clear();	cin>>n;
		for(i = N-1; i; i--) {
			if(f[i]<=n) {
				G.push_back(i);
				n -= f[i];
			}
		}
		reverse(G.begin(), G.end());
		dp[0][0] = 1;
		dp[0][1] = (G[0]-1)>>1;
		for(ll i = 1; i < G.size(); i++) {
			dp[i][0] = dp[i-1][0] + dp[i-1][1];
			dp[i][1] = dp[i-1][0] * ((G[i]-G[i-1]-1)/2) + dp[i-1][1] * ((G[i]-G[i-1])/2);
		}
		cout<<dp[G.size()-1][0]+dp[G.size()-1][1]<<endl;
	}
	return 0;
}