HDU 4099 Revenge of Fibonacci

字典树。刚开始用java乱搞了一发，TLE。然后考虑怎样才能在一堆字符串中找到某个串，使得目标串为它的前缀（或者找不到）。也只能想到字典树了。先把每次的数列求出来，多保留几位来保证精度。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", n)
#define PC(n) printf("Case #%d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 9901;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 100000;
const int M = 66;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

struct node
{
    node *next[10];
    int num;
    node(){};
    node(int a){num = a;CLR(next, 0);};
}root;

int a[M], b[M], c[M];
char s[M];

void insert(node *p, int l, int k)
{
    FD(i, l, max(0, l - 40))
    {
        if(p -> next[c[i]])p = p -> next[c[i]];
        else
        {
            node *t = new node(k);
            p -> next[c[i]] = t;
            p = t;
        }
    }
}

int find(node *p)
{
    int l = strlen(s), k;
    REP(i, l)
    {
        k = s[i] - '0';
        if(p -> next[k])p = p -> next[k];
        else return -1;
    }
    return p -> num;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    root = node(0);
    int t, l;
    a[0] = b[0] = c[0] = 1;
    insert(&root, 1, 0);
    FF(i, 2, N)
    {
        t = 0;
        REP(j, M)
        {
            c[j] = a[j] + b[j] + t;
            t = c[j] / 10;
            c[j] %= 10;
        }
        l = M;
        while(!c[l - 1])l--;
        insert(&root, l, i);
        if(l > 55)REP(j, M - 1)c[j] = c[j + 1], b[j] = b[j + 1];
        REP(j, l)a[j] = b[j], b[j] = c[j];
    }
    RI(t);
    FFF(cas, 1, t)
    {
        RS(s);
        PC(cas);
        PI(find(&root));
    }
    return 0;
}