题意：第数n是斐波那契的第k个数，让你求有多少满足这个的序列（成员非负，不下降）

思路：

1          2          3           4         5           6         7
0          1          1           2         3           5         8
fib(1)    fib(2)    fib(3)  fib(4)     fib(5)      fib(6)    fib(7)
0a+1b  0a+1b   1a+1b   1a+2b      2a+3b       3a+5b     5a+8b
a的系数是fib(i-1),b的系数是fib(i)

代码：

#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<map>
#include<stack>
#include<set>
#include<queue>#define maxn 500
//#define ll long long
#define int long longint n;
int a, b;
int minn = 0, maxx = 0x3f3f3f3f3f;
int x, y;
using namespace std;
int f(int x)
{if (x == 1)return 0;if (x == 2)return 1;return f(x - 1) + f(x - 2);
}
void solve()
{int n, k;cin >> n >> k;int a=0, b=1, c=0;//for (int i = 1; i <= 100; i++)//我们会发现第二十九个就超出了2*10^5即给的n最大值。//	////{//	if (f(i) > 200000)//	{//		cout << "     " << i << endl;//		break;//	}//}if (k > 29)//注意特判{cout << 0 << endl;return;}for (int i = 1; i <= k; i++)//fib{if (i == 1){c = 0;continue;}if (i == 2){c=1;continue;}c = a + b;a = b;b = c;}int ans = 0;//a*x+b*y=n->y=(n-ax)/b&&intfor (int i = 0; i <= n/2; i++){if ((n - a * i) % b == 0 && ((n - a * i) / b) >= i)ans++;}cout << ans<<endl;
}
signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while(t--)solve();return 0;
}