UVa 12333 - Revenge of Fibonacci （大数 + 字典树）

挺简单的一道字典树，注意计算范围， 如果序号大于等于100000就返回-1。

套用的刘汝佳的大数加法。

#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;

struct BigInteger {
  static const int BASE = 100000000;
  static const int WIDTH = 8;
  vector<int> s;

  BigInteger(long long num = 0) { *this = num; } // 构造函数
  BigInteger operator = (long long num) { // 赋值运算符
    s.clear();
    do {
      s.push_back(num % BASE);
      num /= BASE;
    } while(num > 0);
    return *this;
  }
  BigInteger operator = (const string& str) { // 赋值运算符
    s.clear();
    int x,len = (str.length() - 1) / WIDTH + 1;
    for(int i = 0; i < len; i++) {
      int end = str.length() - i*WIDTH;
      int start = max(0,end - WIDTH);
      sscanf(str.substr(start,end-start).c_str(),"%d",&x);
      s.push_back(x);
    }
    return *this;
  }
  BigInteger operator + (const BigInteger& b) const {
    BigInteger c;
    c.s.clear();
    for(size_t i = 0,g = 0; ; i++) {
      if(g == 0 && i >= s.size() && i >= b.s.size()) break;
      int x = g;
      if(i < s.size()) x += s[i];
      if(i < b.s.size()) x += b.s[i];
      c.s.push_back(x % BASE);
      g = x / BASE;
    }
    return c;
  }
};

ostream& operator << (ostream &out,const BigInteger& x) {
  out << x.s.back();
  for(int i = x.s.size()-2; i >= 0; i--) {
    char buf[20];
    sprintf(buf,"%08d",x.s[i]);
    for(size_t j = 0; j < strlen(buf); j++) out << buf[j];
  }
  return out;
}

istream& operator >> (istream &in,BigInteger& x) {
  string s;
  if(!(in >> s)) return in;
  x = s;
  return in;
}

const int maxn = 10 + 7;
const int maxv = 100000;

struct DicTree{
    int k;
    DicTree* next[maxn];
};

DicTree *newtree (int id){
    DicTree *p;
    p = new DicTree;
    p -> k = id;
    for(int i = 0; i < 10; i++) {
        p -> next[i] = NULL;
    }
    return p;
}

DicTree *root;

void creat(BigInteger x,int id) {
    int t,tt,d;
    DicTree *p = root;
    for(int i = (int) x.s.size() - 1; i >= max(0,(int) x.s.size() - 10); --i) {
        if(i == (int) x.s.size() - 1) {
            d = 1;
            t = log10(x.s[i]);
            tt = x.s[i];
            for(int j = 0; j < t; ++j) {
                d *= 10;
            }
            while(d) {
                if(p -> next[tt / d]) {
                    p = p -> next[tt / d];
                } else {
                    p -> next[tt / d] = newtree(id);
                    p = p -> next[tt / d];
                }
                tt %= d;
                d /= 10;
            }
        } else {
            d = 10000000;
            tt = x.s[i];
            while(d) {
                if(p -> next[tt / d]) {
                    p = p -> next[tt / d];
                } else {
                    p -> next[tt / d] = newtree(id);
                    p = p -> next[tt / d];
                }
                tt %= d;
                d /= 10;
            }
        }
    }
}

int findtree(string s) {
    DicTree *p = root;
    for(size_t i = 0; i < s.size(); i++) {
        if(p -> next[s[i] - '0']) {
            p = p -> next[s[i] - '0'];
        } else {
            return -1;
        }
    }
    return p -> k;
}

int main() {
    int T,Case(0);
    string x;
    BigInteger a(1),b(1),c(1);
    root = new DicTree;
    for(int i = 0; i < 10; i++) {
        root -> next[i] = NULL;
    }
    root -> next[1] = newtree(0);
    for(int i = 2; i < maxv; i++) {
        c = a + b;
        creat(c,i);
        b = a;
        a = c;
    }
    cin >> T;
    while(T--) {
        x.clear();
        cin >> x;
        cout << "Case #" << ++Case << ": " << findtree(x) << endl;
    }
    return 0;
}