[斐波那契拆分 乱搞 数学] 51Nod 1350 斐波那契表示

可以证明每次选最大的fib数减去一定是最优解之一

然后预处理G[fib[i]-1] 

额 代码横跨noip前后 意识模糊 一片mess

#include<cstdio>#include<cstdlib>#include<algorithm>#include<map>using namespace std;typedef long long ll;ll fib[205]; int maxn;ll F[205];inline ll Sum(ll n){  if (n==0) return 0;  if (n==1) return 1;  if (n==2) return 2;  ll ret=0;  int pos=upper_bound(fib,fib+maxn+1,n)-fib-1;  ret=F[pos]+Sum(n-fib[pos])+n-fib[pos]+1;  return ret;}int main(){  ll T,n;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  fib[0]=0; fib[1]=1;  F[2]=0; F[3]=1; F[4]=2;  for (int i=2;;i++){    fib[i]=fib[i-1]+fib[i-2];    if (fib[i]-1>2)      F[i]=F[i-1]+F[i-2]+fib[i-2];    maxn=i;    if (fib[i]>1e17) break;  }  scanf("%lld",&T);  while (T--)    scanf("%lld",&n),printf("%lld\n",Sum(n));  return 0;}