标签：acm   hdu   

第一题：

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Description

Neal is very curious about combinatorial problems, and now here comes a problem about words. Knowing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie.

Since Jiejie can‘t remember numbers clearly, he just uses sticks to help himself. Allowing for Jiejie‘s only 20071027 sticks, he can only record the remainders of the numbers divided by total amount of sticks.

The problem is as follows: a word needs to be divided into small pieces in such a way that each piece is from some given set of words. Given a word and the set of words, Jiejie should calculate the number of ways the given word can be divided, using the words in the set.

abcd 
4 
a 
b 
cd 
ab

Case 1: 2

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#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
#define maxn 400000 + 10
#define M 300000 + 10
#define mod 20071027

int tree[maxn][27], val[maxn], top;
int d[maxn];
char s[M], tmp[105];

void init()
{
    top = 1;
    memset(tree[0], 0, sizeof tree[0]);
    memset(d, 0, sizeof d);
}

void insert(char *s, int v)
{
    int len = strlen(s), u = 0;
    for(int i = 0; i < len; i++)
    {
        int c = s[i] - 'a' ;
        if(!tree[u][c])
        {
            memset(tree[top], 0, sizeof tree[top]);
            val[top] = 0;
            tree[u][c] = top++;
        }
        u = tree[u][c];
    }
    val[u] = v;
}

int len ;

int query(char *s, int pos)
{
    int u = 0;
    int res = 0;
    for(int i = pos; i < len; i++)
    {
        int c = s[i] - 'a' ;
        u = tree[u][c];
        if(!u) return res;
        if(val[u])
        {
            res = (res + d[i + 1] ) % mod ;
        }
    }
    return res;
}

int main()
{
    int kase = 0;
    while(~scanf("%s", s))
    {
        init();
        len = strlen(s);
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", tmp);
            insert(tmp, 1);
        }
        d[len] = 1;
        for(int i = len - 1; i >= 0; i--)
        {
            d[i] = query(s, i);
        }
        printf("Case %d: %d\n", ++kase, d[0]);
    }
    return 0;
}

ZYB loves Xor I

2
5
4 0 2 7 0
5
2 6 5 4 0

Case #1: 36
Case #2: 40

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MOD 998244353
#define M 2000010
using namespace std;

int top;// top 表示节点个数
int tree[M][2]; // 字典树
int val[M]; //字符串的权值 当val[i] 大于0时i是单词节点

void init()
{
    top=1;
    //memset(tree, 0, sizeof tree);
    memset(tree[0], 0, sizeof tree[0]);
    memset(val, 0, sizeof val);
}

void insert(int v)
{
    int u = 0;//根节点
    for(int i = 0; i < 30 ; i++)
    {
        int t = v & ( 1<< i );
        if(t) t = 1;
        if(!tree[u][t])
        {
            memset(tree[top], 0, sizeof tree[top]);
            val[top] = 1;
            tree[u][t] = top++;
        }
        else
        {
            val[tree[u][t]]++;
        }
        u = tree[u][t];
    }
}

int ans;
void find(int v)
{
    int u = 0;
    for(int i = 0; i < 30; i++)
    {
        int t = v & (1 << i);
        if(t) t = 1;
        if(tree[u][t ^ 1])
        ans = ( ans + (val[tree[u][t ^ 1]]) * (1 << i)) % MOD;
        u = tree[u][t] ;
    }
}

int a[M];

int main()
{
   int T, n;
   scanf("%d", &T);
   int kase = 0;
   while(T--)
   {
       init();
       ans = 0;
       scanf("%d", &n);
       for(int i=0; i<n; i++)
        {
            scanf("%d", &a[i]);
            insert(a[i]);
        }
       for(int i=0; i<n; i++)
        find(a[i]);
       printf("Case #%d: %d\n", ++kase, ans);
   }
    return 0;
}

字典树数组形式写法

标签：acm   hdu   

原文地址：http://blog.csdn.net/dojintian/article/details/47300183