1 100 -1
12 11
把斐波那契数列插入ac自动机，然后数位dp进行二分查找，
代码：
/* ***********************************************
Author :rabbit
Created Time :2014/8/15 16:27:37
File Name :3.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
struct Trie{
ll next[1010][10],end[1010],fail[1010];
ll root,L;
void init(){
L=0;
root=newnode();
}
ll newnode(){
for(ll i=0;i<10;i++)
next[L][i]=-1;
end[L++]=0;
return L-1;
}
void insert(ll x){
ll a[15],len=0;
while(x){
a[len++]=x%10;
x/=10;
}
//for(ll i=len-1;i>=0;i--)cout<<a[i]<<" ";cout<<endl;
ll now=root;
for(ll i=len-1;i>=0;i--){
if(next[now][a[i]]==-1)
next[now][a[i]]=newnode();
now=next[now][a[i]];
}
end[now]=1;
}
void build(){
queue<ll> Q;
fail[root]=root;
for(ll i=0;i<10;i++)
if(next[root][i]==-1)
next[root][i]=root;
else{
fail[next[root][i]]=root;
Q.push(next[root][i]);
}
while(!Q.empty()){
ll now=Q.front();
Q.pop();
end[now]|=end[fail[now]];
for(ll i=0;i<10;i++)
if(next[now][i]==-1)
next[now][i]=next[fail[now]][i];
else{
fail[next[now][i]]=next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
}ac;
ll fib[60],dp[15][1010],seq[60],POW[20];
ll suf[15],num[15];
ll dfs(ll pos,ll st,bool flag){
if(pos==0)return 0;
if(flag&&dp[pos][st]!=-1)return dp[pos][st];
ll u=flag?9:num[pos];
ll ans=0;
for(ll i=0;i<=u;i++){
ll state=ac.next[st][i];
if(ac.end[state]){
if(flag||i<u)ans+=POW[pos-1];//后面的任意取
else ans+=suf[pos-1]+1;
}
else ans+=dfs(pos-1,state,flag||i<u);
}
if(flag)dp[pos][st]=ans;
return ans;
}
ll cal(ll x){
ll len=0;
suf[0]=0;
//cout<<"x="<<x<<endl;
while(x){
num[++len]=x%10;
suf[len]=suf[len-1]+x%10*POW[len-1];
x/=10;
}
//cout<<"suf: ";for(ll i=1;i<=len;i++)cout<<suf[i]<<" ";cout<<endl;
return dfs(len,ac.root,0);
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data1.out","w",stdout);
fib[1]=1;
fib[2]=1;
for(ll i=3;i<=60;i++)
fib[i]=fib[i-1]+fib[i-2];
memset(dp,-1,sizeof(dp));
ac.init();
for(ll i=1;i<=56;i++)
if(fib[i]>10)ac.insert(fib[i]);
POW[0]=1;
for(ll i=1;i<=15;i++)POW[i]=10*POW[i-1];
ac.build();
for(ll i=2;i<=56;i++){
ll l=13,r=POW[12];
while(l<r){
ll mid=(l+r)/2;
ll ret=cal(mid);
// cout<<"han "<<i<<" "<<l<<" "<<r<<" "<<ret<<endl;
if(ret<fib[i])l=mid+1;
else r=mid;
}
seq[i]=l;
}
//for(ll i=2;i<=56;i++)cout<<seq[i]<<endl;return 0;
ll n;
while(cin>>n&&n!=-1){
ll ans=1e15;
for(ll i=2;i<=56;i++){
ll tt=n-seq[i];
if(tt<0)tt=-tt;
ans=min(ans,tt);
}
cout<<ans<<endl;
}
return 0;
}
HDU 4518 ac自动机+数位dp,布布扣,bubuko.com
原文：http://blog.csdn.net/xianxingwuguan1/article/details/38588757