hdu4099(Trie)

题意：

　　给定一个长度不超过40的数字串s，求斐波那契数列的前十万项中，最小的一个前缀为s的数的下标。

解决：

　　高精度算出前十万个斐波那契数，每个取前40位，建trie，查询O(40)

（本地建立Trie跑了20s，让我知道了为啥服务器那么贵）

#include <bits/stdc++.h>

int a[22000], b[22000], c[22000];
int l1, l2, l3;
int op[55], len;

struct Node{
    Node *ch[10];
    int ans;
    Node()
    {
        for (int i = 0; i < 10; ++i)
            ch[i] = NULL;
        ans = -1;
    }
};

struct Trie{
    Node *root;
    void init()
    {
        root = new Node();
    }
    void trieInsert(int index)
    {
        Node *p = root;
        for (int i = l3; i >= std::max(l3-39, 1); --i) {
            if ( p -> ch[c[i]] == NULL) {
                p -> ch[c[i]] = new Node();
                p-> ch[c[i]] -> ans = index;
            }
            p = p ->ch[c[i]];
        }
        p -> ans = index;
    }
    int trieQuery()
    {
        Node *p = root;
        for (int i = 1; i <= len; ++i) {
            if ( p -> ch[op[i]] == NULL)
                return -1;
            p = p -> ch[op[i]];
        }
        return p -> ans;
    }
}tr;

void add()
{
    int carry = 0;
    for (int i = 1; i <= std::max(l1, l2); ++i) {
        c[i] = a[i] + b[i] + carry;
        if (c[i] >= 10) {
            carry = 1;
            c[i] -= 10;
        }
        else
            carry = 0;
    }
    if (carry == 1) {
        c[++l3] = 1;
    }
}

void init()
{
    tr.init();
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    memset(c, 0, sizeof c);
    a[++l1] = 1;
    b[++l2] = 1;
    c[++l3] = 1;
    tr.trieInsert(0);
    for (int i = 2; i < 100000; ++i) {
        add();
        tr.trieInsert(i);
        for (int j = 1; j <= l2; ++j) {
            a[j] = b[j];
        }
        for (int j = 1; j <= l3; ++j) {
            b[j] = c[j];
        }
        l1 = l2;
        l2 = l3;
    }
}

char buf[55];

int main()
{
    init();
    int T;
    scanf("%d", &T);
    for (int t = 1; t <= T; ++t) {
        scanf("%s", buf+1);
        len = strlen(buf+1);
        for (int i = 1; i <= len; ++i) {
            op[i] = buf[i] - '0';
        }
        printf("Case #%d: %d
", t, tr.trieQuery());
    }
}