习题5-15(uva-12333)-优快云博客

本文链接：https://blog.youkuaiyun.com/seanbill/article/details/109575420

本文介绍了一种使用C++实现的大数加法运算，并结合字典树（Trie）进行快速查找的方法。该方法适用于需要处理大量大数运算及检索的应用场景，如密码学中的密钥生成等。

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#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>

#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int BigSize = 50;
const int FSIZE = 1e5 + 11;
const int TSIZE = 1e7;
//为什么只对前50位进行运算可以保证精度 这我想不到
struct BigInt {
	int arr[BigSize + 1];
	BigInt() {
		memset(arr, 0, sizeof(arr));
	}
	void Moved() {
		for (int i = 0; i < BigSize; ++i) {
			arr[i] = arr[i + 1];
		}
		arr[BigSize] = 0;
	}
	bool IsFilled() {
		return arr[BigSize];
	}
	BigInt operator+(BigInt& a) {
		int carry = 0,sum = 0;

		BigInt ans;
		if (IsFilled() || a.IsFilled()) { Moved(); a.Moved(); }

		for (int i = 0; i <= BigSize; ++i) {
			sum = arr[i] + a.arr[i] + carry;
			ans.arr[i] = sum % 10;
			carry = sum / 10;
		}
		return ans;
	}
}f[FSIZE];

struct Tree {
	Tree(){
		memset(this, 0, sizeof(this));
	}
	int son[10];
	int index;

}tree[TSIZE];

int treeCnt = 1;

void Insert(BigInt& in,int icnt) {
	int cur = 1,index = BigSize,id,isnew = 0;
	while (index >= 0 && in.arr[index] == 0) index--;

	for (; index >= 0; index--) {
		id = in.arr[index];
		if (!tree[cur].son[id]) { isnew = 1; tree[cur].son[id] = ++treeCnt; }
		cur = tree[cur].son[id];
		if (isnew) tree[cur].index = icnt;
	}
}
int Query(string &in) {
	int ret = -1,id = 0,cur = 1;
	for (int i = 0; i < in.size(); ++i) {
		id = in[i] - '0';
		if (!tree[cur].son[id]) return -1;
		cur = tree[cur].son[id];
		ret = tree[cur].index;
	}
	return ret;
}

int main()
{
	f[0].arr[0] = f[1].arr[0] = 1;
	Insert(f[0], 0);
	for (int i = 2; i < 100000; ++i) f[i] = f[i - 1] + f[i - 2], Insert(f[i], i);

	string temp;
	int n,m = 0;
	cin >> n;
	while (n--) {
		cin >> temp;
		cout <<"Case #"<<++m<<": " << Query(temp) << endl;
	}
	return 0;
}