由于是斐波那契数列，所以$x_i+x_j<=x_k,i<j<k$
所以猜测可以贪心选择两边近的数处理。
1 #include<cstdio> 2 #include<algorithm> 3 #define ll long long 4 #define mid (l+r>>1) 5 using namespace std; 6 ll f[505],tot=1; 7 inline ll findl(ll x) 8 { 9 int l=1,r=tot,ans=1; 10 while(l<=r) 11 { 12 if(f[mid]<=x)ans=mid,l=mid+1; 13 else r=mid-1; 14 } 15 return f[ans]; 16 } 17 inline ll findr(ll x) 18 { 19 int l=1,r=tot,ans=1; 20 while(l<=r) 21 { 22 if(f[mid]>=x)ans=mid,r=mid-1; 23 else l=mid+1; 24 } 25 return f[ans]; 26 } 27 int solve(ll x) 28 { 29 ll l=findl(x),r=findr(x); 30 if(l==r)return 1; 31 if(x-l<=r-x)return solve(x-l)+1; 32 return solve(r-x)+1; 33 } 34 int main() 35 { 36 f[0]=f[1]=1; 37 while(f[tot-1]<=4e17) 38 f[++tot]=f[tot-1]+f[tot-2]; 39 int t;scanf("%d",&t); 40 ll x; 41 while(t--)scanf("%lld",&x), 42 printf("%d ",solve(x)); 43 }