UVa-12333：Revenge of Fibonacci 高精度

之前自己仿照紫书上写了高精度库，完善了乘法、减法，并且通过了和C++高精度库GMP的对拍测试，并在一些OJ上过了一些高精度的模板题，代码仓库地址：https://github.com/Edward-Elric233/BigInt

求解思路

题目的意思是求前100000斐波那契数列中某个前缀（不超过40个字符）第一次出现的位置。刚开始我的想法很简单，先求出这十万个斐波那契数列的前缀，然后每次读入的时候查找一遍就可以了，结果超时了。每次查找的复杂度是O(1e5∗40)O(1e5 * 40)O(1e5∗40)，有5e4个样例，这样10s也会超时。

但是我发现这个数据结构只有查找操作，因此使用unordered_map再合适不过了。我将所有前缀保存在这个unordered_map中，每次查找近似都是O(1)O(1)O(1)的，因此肯定不会超时。

需要注意的是这里将前缀保存的时候需要将每个数字的所有前缀都保存，例如对于123，就要保存前缀1，12，123。

AC代码

// Copyright(C), Edward-Elric233
// Author: Edward-Elric233
// Version: 1.0
// Date: 2021/8/8
// Description:#include <string>
#include <vector>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <unordered_map>class BigInt {public:using ll = long long;static constexpr int BASE = 1e8;    //基数是1e8，即满1e8才进一位static constexpr int WIDTH = 8;     //每一位的十进制长度是8位，主要用于字符串和转换std::vector<int> bit;               //保存BigInt的每一位，小端模式：低位低字节，高位高字节bool negative;                      //记录数字是否是负数static void trim(std::string &s);                           //trim函数：删除头部和尾部的空格static void num2big(std::vector<int> &bit, ll num);         //实际处理函数：将正数num放入bit数组中static void str2big(std::vector<int> &bit, std::string &s); //实际处理函数：将正数字符串放入bit数组中static bool less(const BigInt &lhs, const BigInt &rhs);     //比较两个数字的绝对值的大小BigInt(ll num = 0);BigInt(std::string s); BigInt & operator =(ll num);         //必须返回BigInt&，与内置类型一致BigInt & operator =(std::string s);BigInt operator -() const;friend std::ostream &operator << (std::ostream &os, const BigInt &bigInt);friend std::istream &operator >> (std::istream &is, BigInt &bigInt);friend BigInt operator + (const BigInt &lhs, const BigInt &rhs);friend BigInt operator - (const BigInt &lhs, const BigInt &rhs);friend BigInt operator * (const BigInt &lhs, const BigInt &rhs);friend BigInt operator / (const BigInt &lhs, const BigInt &rhs);friend bool operator < (const BigInt &lhs, const BigInt &rhs);friend bool operator > (const BigInt &lhs, const BigInt &rhs);friend bool operator == (const BigInt &lhs, const BigInt &rhs);friend bool operator != (const BigInt &lhs, const BigInt &rhs);friend bool operator <= (const BigInt &lhs, const BigInt &rhs);friend bool operator >= (const BigInt &lhs, const BigInt &rhs);BigInt & operator += (const BigInt &rhs);BigInt & operator -= (const BigInt &rhs);BigInt & operator *= (const BigInt &rhs);BigInt & operator /= (const BigInt &rhs);std::string toString() const;BigInt & setOppo();                 //取相反数bool isNegative() const;            //判断是否是一个负数
};std::ostream & operator << (std::ostream &os, const BigInt &bigInt);
std::istream & operator >> (std::istream &is, BigInt &bigInt);
BigInt operator + (const BigInt &lhs, const BigInt &rhs);
BigInt operator - (const BigInt &lhs, const BigInt &rhs);
BigInt operator * (const BigInt &lhs, const BigInt &rhs);
BigInt operator / (const BigInt &lhs, const BigInt &rhs);
bool operator < (const BigInt &lhs, const BigInt &rhs);
bool operator > (const BigInt &lhs, const BigInt &rhs);
bool operator == (const BigInt &lhs, const BigInt &rhs);
bool operator != (const BigInt &lhs, const BigInt &rhs);
bool operator <= (const BigInt &lhs, const BigInt &rhs);
bool operator >= (const BigInt &lhs, const BigInt &rhs);BigInt BigInt::operator-() const {BigInt ret = *this;ret.negative = !ret.negative;return ret;
}BigInt & BigInt::setOppo() {negative = !negative;return *this;
}bool BigInt::isNegative() const {return negative;
}void BigInt::num2big(std::vector<int> &bit, ll num) {ll x;//必须使用do while，循环至少应该执行一遍，处理num是0的情况do {x = num % BASE;bit.push_back(static_cast<int>(x));num /= BASE;} while (num);
}BigInt::BigInt(ll num):negative(false) {if (num < 0) {num = -num;negative = true;}num2big(this->bit, num);
}void BigInt::str2big(std::vector<int> &bit, std::string &s) {int len = (s.size() - 1) / WIDTH + 1;       //ceil得到lenint start, end;for (int i = 0; i < len; ++i) {end = s.size() - i * WIDTH;start = std::max(0, end - WIDTH);bit.push_back(std::stoi(s.substr(start, end - start)));}
}BigInt::BigInt(std::string s):negative(false) {trim(s);if (s[0] == '-') {negative = true;s.erase(s.begin());}str2big(this->bit, s);
}void BigInt::trim(std::string &s) {auto ltrim = [](std::string &s) {s.erase(s.begin(), std::find_if(s.begin(), s.end(), [](unsigned char c) -> bool {//找到第一个非空字符return !std::isspace(c);}));};auto rtrim = [](std::string &s) {s.erase(std::find_if(s.rbegin(), s.rend(), [](unsigned char c) -> bool {//找到最后一个非空字符return !std::isspace(c);}).base(), s.end());};ltrim(s);rtrim(s);
}BigInt &BigInt::operator=(ll num) {bit.clear();negative = false;if (num < 0) {num = -num;negative = true;}num2big(this->bit, num);return *this;
}BigInt &BigInt::operator=(std::string s) {bit.clear();negative = false;trim(s);if (s[0] == '-') {negative = true;s.erase(s.begin());}str2big(this->bit, s);return *this;
}std::ostream &operator << (std::ostream &os, const BigInt &bigInt) {if (bigInt.negative) {//输出负号os << "-";}auto &bit = bigInt.bit;//以大端字节序输出os << bit.back();//除了最后一位，其他的如果有前导0不能忽略os << std::setfill('0');for (int i = bit.size() - 2; i >= 0; --i) {os << std::setw(BigInt::WIDTH) << bit[i];}os << std::setfill(' ');return os;
}std::istream &operator >> (std::istream &is, BigInt &bigInt) {std::string s;if (is >> s) {//必须判断是否读入成功bigInt = s;}return is;
}BigInt & BigInt::operator+=(const BigInt &rhs) {if (!negative && rhs.negative) {BigInt tmp = rhs;return *this = (tmp -= this->setOppo());} else if (negative && !rhs.negative) {return *this -= -rhs;}auto &rbit = rhs.bit;int max_len = std::max(bit.size(), rbit.size());int min_len = std::min(bit.size(), rbit.size());int g = 0;      //进位for (int i = 0; i < min_len; ++i) {bit[i] += rbit[i] + g;g = bit[i] / BASE;bit[i] %= BASE;}if (bit.size() < rbit.size()) {for (int i = min_len; i < max_len; ++i) {bit.push_back(g + rbit[i]);g = bit.back() / BASE;bit.back() %= BASE;}} else {for (int i = min_len; i < max_len; ++i) {bit[i] += g;g = bit[i] / BASE;bit[i] %= BASE;if (!g) break;}}if (g) {//加法的进位最多为1bit.push_back(g);}return *this;
}BigInt operator + (const BigInt &lhs, const BigInt &rhs) {BigInt ret = lhs;ret += rhs;return std::move(ret);
}BigInt & BigInt::operator-=(const BigInt &rhs) {if (!negative && !rhs.negative) {if (*this >= rhs) {} else {BigInt tmp = rhs;tmp -= *this;return *this = tmp.setOppo();}} else if (!negative && rhs.negative) {this->setOppo();*this += rhs;this->setOppo();return *this;} else if (negative && !rhs.negative) {return *this += -rhs;} else {BigInt tmp = -rhs;this->setOppo();return *this = (tmp -= *this);}auto &rbit = rhs.bit;for (int i = 0; i < rbit.size(); ++i) {if (bit[i] < rbit[i]) {bit[i] += BASE;bit[i + 1] -= 1;}bit[i] -= rbit[i];}for (int i = rbit.size(); i < bit.size(); ++i) {if (bit[i] >= 0) {break;}bit[i] += BASE;bit[i + 1] -= 1;}//删除前导0for (int i = bit.size() - 1; i > 0; --i) {if (!bit[i]) bit.pop_back();}return *this;
}BigInt operator - (const BigInt &lhs, const BigInt &rhs) {BigInt res = lhs;res -= rhs;return std::move(res);
}BigInt & BigInt::operator*=(const BigInt &rhs) {//负负得正if (negative && rhs.negative || !negative && !rhs.negative) {negative = false;} else {negative = true;}auto &rbit = rhs.bit;constexpr ll LBASE = BASE;std::vector<ll> c(bit.size() + rbit.size(), 0);for (int i = 0; i < bit.size(); ++i) {for (int j = 0; j < rbit.size(); ++j) {c[i + j] += static_cast<ll>(bit[i]) * static_cast<ll>(rbit[j]);//在这里处理进位防止溢出if (c[i + j] >= LBASE) {//有必要再进行除法，毕竟除法比较慢c[i + j + 1] += c[i + j] / LBASE;c[i + j] %= LBASE;}}}//处理进位for (int i = 0; i < c.size(); ++i) {if (c[i] >= LBASE) {//有必要再进行除法，毕竟除法比较慢c[i + 1] += c[i] / LBASE;c[i] %= LBASE;}}//删除前导0for (int i = c.size() - 1; i > 0; --i) {    //至少留一位if (!c[i]) c.pop_back();else break;}bit.resize(c.size());for (int i = 0; i < c.size(); ++i) {bit[i] = static_cast<int>(c[i]);}return *this;
}BigInt operator * (const BigInt &lhs, const BigInt &rhs) {BigInt res = lhs;res *= rhs;return std::move(res);
}std::string BigInt::toString() const {std::ostringstream os;os << *this;return os.str();
}bool BigInt::less(const BigInt &lhs, const BigInt &rhs) {if (lhs.bit.size() != rhs.bit.size()) return lhs.bit.size() < rhs.bit.size();for (int i = lhs.bit.size() - 1; i >= 0; --i) {if (lhs.bit[i] != rhs.bit[i]) return lhs.bit[i] < rhs.bit[i];}//相等return false;
}bool operator < (const BigInt &lhs, const BigInt &rhs) {if (!lhs.negative && !rhs.negative) {return BigInt::less(lhs, rhs);} else if (lhs.negative && !rhs.negative) {return true;} else if (!lhs.negative && rhs.negative) {return false;} else if (lhs.negative && rhs.negative) {//都是负数if (BigInt::less(lhs, rhs)) {return false;} else if (BigInt::less(rhs, lhs)) {return true;} else {return false;}}
}bool operator > (const BigInt &lhs, const BigInt &rhs) {return rhs < lhs;
}bool operator == (const BigInt &lhs, const BigInt &rhs) {return !(lhs < rhs) && !(rhs < lhs);
}
bool operator != (const BigInt &lhs, const BigInt &rhs) {return (lhs < rhs) || (rhs < lhs);
}bool operator <= (const BigInt &lhs, const BigInt &rhs) {return !(rhs < lhs);
}bool operator >= (const BigInt &lhs, const BigInt &rhs) {return !(lhs < rhs);
}using namespace std;constexpr int MAXN = 1e5;
vector<BigInt> fib;
unordered_map<string, int> str_hash;int main() {ios::sync_with_stdio(false);fib.push_back(1);fib.push_back(1);for (int i = 2; i < MAXN; ++i) {fib.push_back(fib[i - 1] + fib[i - 2]);}for (int i = 0; i < MAXN; ++i) {string line;auto &bit = fib[i].bit;line.append(to_string(bit.back()));for (int i = bit.size() - 2, j = std::max(0, int(bit.size()) - 6); i >= j; --i) {const string &number = to_string(bit[i]);for (int i = 0, j = 8 - number.size(); i < j; ++i) line.append("0");line.append(number);}
//        cout << line << endl;for (int ii = 1, jj = std::min(int(line.size()), 40); ii <= jj; ++ii) {const string &w = line.substr(0, ii);
//            cout << w << endl;if (str_hash.count(w)) {if (i < str_hash[w]) {str_hash[w] = i;}} else {str_hash[w] = i;}}}
//    for (int i = 0; i < 10; ++i) cout << fibHead[i] << " ";
//    cout << "\n";int T;cin >> T;string line;for (int caseI = 1; caseI <= T; ++caseI) {cin >> line;cout << "Case #" << caseI << ": ";if (str_hash.count(line)) {cout << str_hash[line] << "\n";} else {cout << "-1\n";}}
}

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UVa-12333：Revenge of Fibonacci 高精度

求解思路

AC代码

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