【百度之星

1001.Problem A

http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=690&pid=1001

　　类似倒水原理，但是这里是连乘，对于连续累积S。 x = S[a...b] = S[b] % 9973 / S[a - 1] % 9973。 也就是求

a * x = b %9973 线性方程组的解。EXGCD拓展GCD求解。数论题。

Promble B

　　斐波那契数列。可以大整数，可以java。我贴下Java代码，好久没用了。

import java.math.*;
import java.io.*;
import java.text.*;
import java.util.Scanner;

public class Main {
    public static void main(String []args){ 
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()){
            BigInteger r = cin.nextBigInteger();
            BigInteger bigTwo = new BigInteger("2");
            BigInteger m = BigInteger.ZERO;
            BigInteger n = BigInteger.ZERO;
            BigInteger ret = BigInteger.ZERO;
            while(m.compareTo(r.divide(bigTwo)) <= 0){
                n = r.subtract(m);
                BigInteger fz = BigInteger.ONE;
                BigInteger fm = BigInteger.ONE;
                BigInteger left = (n.subtract(m)).add(BigInteger.ONE);
                BigInteger right = n;
                while(left.compareTo(right) <= 0){
                    fz = fz.multiply(left);
                    left = left.add(BigInteger.ONE);
                }
                left = m;
                while(left.compareTo(BigInteger.ONE) >= 0){
                    fm = fm.multiply(left);
                    left = left.subtract(BigInteger.ONE);
                }
                ret = ret.add(fz.divide(fm));
                m = m.add(BigInteger.ONE);
            }
            System.out.println(ret);
        }
    }
}

Problem C

　　字典树。贴一下自己敲的Trie，下次没准用得上。

#include <iostream>
#include <cmath>
#include <cstring>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

const int Max = 26;
class TrieNode {
public:
    int num;
    TrieNode *next[Max];
    TrieNode() {
        memset(next, NULL, sizeof next);
        num = 0;
    }
};
class Trie {
public:
    
    Trie() {
        root = new TrieNode();
    }
    ~Trie(){
        free(root);
    }
    // Inserts a word into the trie.
    void insert(char* word) {
        TrieNode* p = root;
        int len = strlen(word);
        for(int i = 0; i < len; i ++){
            int id = word[i] - 'a';
            if(p->next[id] == NULL)
                p->next[id] = new TrieNode();
            p = p->next[id];
            p->num ++;
        }
    }
    // Returns if the word is in the trie.
    TrieNode* search(char* word) {
        TrieNode *p = root;
        int len = strlen(word);
        for(int i = 0; i < len; i ++){
            int id = word[i] - 'a';
            if(p->next[id] == NULL) return NULL;
            else{
                p = p->next[id];
            }
        }
        return p;
    }

    void destoryWord(TrieNode* p)
    {
        if(!p) return ;
        for(int i = 0; i < Max; i ++ ){
            destoryWord(p -> next[i]);
        }
        free(p);
    }

    void destorySon(TrieNode * p)
    {
        if(!p) return;
        for(int i = 0; i < Max; i ++ ){
            destoryWord(p -> next[i]);
            p->next[i] = NULL;
        }
    }

    void deleteWord(char* word)
    {
        TrieNode *cur = search(word);
        if(!cur) return ;
        int count = cur -> num;
        destorySon(cur);
        TrieNode *p = root;
        int len = strlen(word);
        for(int i = 0; i < len; i ++){
            int id = word[i] - 'a';
            p = p -> next[id];
            p -> num -= count;
        }
    }

private:
    TrieNode* root;
};

int main()
{
    int n;
    Trie trie;
    scanf("%d",&n);
    getchar();
    char op[10], str[50];
    while(n -- ){
        scanf("%s%s",op, str);
        if(op[0] == 'i'){
            trie.insert(str);
        }else if(op[0] == 'd'){
            trie.deleteWord(str);
        }else {
            TrieNode *rec = trie.search(str);
            if(rec && rec -> num > 0) printf("Yes
");
            else printf("No
");
        }
    }system("pause");
    return 0;
}
/*
11
insert aaaa
insert aaab
insert aab
insert aac
insert ab
search aaa
delete aaa
search aaa
search aaab
search aa
search ab
*/

1004.Problem D

　　字符串hash。一开始用Map<string, int>tle。我还以为是string输入效率不行，改成字典树用char* 输入过得，但之后有看到人家用map哈希过得，我也是很郁闷啊。