UVA12333--Revenge of Fibonacci

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题意：
给你一个数字串，判断他是哪一个Fib数的前缀，有多种答案输出最小的，不存在输出-1。

思路：
大数加法计算斐波那契数，字典树搜索
计算过程中进行字典树的创建，计算时，当斐波那契数大于50位时，只计算高50位即可。

代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define maxn 100100using namespace std;typedef struct Node {    int value;    Node* next[10];    Node() {        value = -1;        for (int i = 0; i < 10; i++)            next[i] = NULL;    }} Node, *Tree;int num[2][maxn];Tree head;void addNode(int *s, int len, int n) {    Node* p = head;    for (int i = 0; i < len; i++) {        int x = s[i];        if (p->next[x] == NULL)            p->next[x] = new Node;        p = p->next[x];        if (p->value == -1)            p->value = n;    }}//计算斐波那契额数并创建字典树void getFibAndTree() {    head = new Node;    head->next[1] = new Node;    head->next[1]->value = 0;    memset(num, 0, sizeof(num));    num[0][0] = num[1][0] = 1;    int b = 0, e = 1, flag = 0;    for (int i = 2; i < 100000; i++) {        int jw = 0;        for (int j = b; j < e; j++) {            int t = num[0][j] + num[1][j] + jw;            jw = 0;            if (t > 9)                jw = 1;            num[flag][j] = t % 10;        }        if (jw)            num[flag][e++] = 1;        if (e - b > 50)            b++;        int len = e - b;        if (len > 40)            len = 40;        int cc[len];        for (int x = 0; x < len; x++)            cc[x] = num[flag][e - 1 - x];        addNode(cc, len, i);        flag = !flag;    }}//搜索int msearch(char *in) {    int len = strlen(in);    Node* p = head;    for (int i = 0; i < len; i++) {        if (p->next[in[i] - '0'] == NULL)            return -1;        p = p->next[in[i] - '0'];    }    return p->value;}int main() {    getFibAndTree();    char in[45];    int n;    cin >> n;    for (int i = 1; i <= n; i++) {        cin >> in;        printf("Case #%d: %d\n", i, msearch(in));    }}

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