BZOJ2796: [Poi2012]Fibonacci Representation

记忆化搜索
所以其实就是乱搜吗….
1017在fib数列里好像就70~80左右的地方，所以数据的数量不多，寻找x的解可以二分找到x大概在fib的哪个位置，然后找x和左右项的差的绝对值y

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#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;void down(int &x,int y){if(x>y)x=y;}const int maxn = 88;map<ll,int>h;ll n;ll f[maxn];int find_(ll x){    int l=2,r=85;    while(l<=r)    {        int mid=(l+r)>>1;        if(x<f[mid]) r=mid-1;        else l=mid+1;    }    return l-1;}ll solve(ll x){    int k=find_(x);    int ans=233;    ll cl=abs(f[k]-x);        if(h.count(cl)>0) ans=h[cl];        else ans=solve(cl);    cl=abs(f[k+1]-x);        if(h.count(cl)>0) down(ans,h[cl]);        else { int l=solve(cl);if(ans>l)ans=l; }    h[x]=ans+1;    return ans+1;}int main(){    h[0]=0;    f[1]=1; for(int i=2;i<maxn;i++) f[i]=f[i-1]+f[i-2],h[f[i]]=1;    int t; scanf("%d",&t);    while(t--)    {        scanf("%lld",&n);        if(h.count(n)>0) printf("%d\n",h[n]);        else printf("%d\n",solve(n));    }    return 0;}