hdu 5207 hash统计

题目大意：求最大的两个数的最大公约数。

思路：数据范围较小，只有10^5，可以考虑哈希。

解法1：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 100001;
int hash_table[N];
int a[N];
int n;

void set( int x )
{
    int i;
    for ( i = 1; i * i < x; i++ )
    {
        if ( x % i == 0 )
        {
            hash_table[i]++;
            hash_table[x / i]++;
        }
    }
    if ( i * i == x ) hash_table[i]++;
}

int solve()
{
    for ( int i = N - 1; i > 0; i-- )
    {
        if ( hash_table[i] >= 2 )
        {
            return i;
        }
    }
}

int main ()
{
    int t;
    scanf("%d", &t);
    for ( int _case = 1; _case <= t; _case++ )
    {
        scanf("%d", &n);
        for ( int i = 0; i < n; i++ )
        {
            scanf("%d", a + i);
        }
        memset( hash_table, 0, sizeof(hash_table) );
        for ( int i = 0; i < n; i++ )
        {
            set(a[i]);
        }
        printf("Case #%d: %d
", _case, solve());
    }
    return 0;
}

复杂度：n ^ 1.5

解法2：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 100001;
int cnt[N];
int mark[N];
int a[N];
int n;

int solve( int r )
{
    for ( int i = r; i > 0; i-- )
    {
        if ( cnt[i] >= 2 ) return i;
    }
}

int main ()
{
    int t;
    scanf("%d", &t);
    for ( int _case = 1; _case <= t; _case++ )
    {
        scanf("%d", &n);
        int maxn = -1;
        memset( mark, 0, sizeof(mark) );
        for ( int i = 0; i < n; i++ )
        {
            scanf("%d", a + i);
            mark[a[i]]++;
            if ( a[i] > maxn ) maxn = a[i];
        }
        memset( cnt, 0, sizeof(cnt) );
        for ( int i = 2; i <= maxn; i++ )
        {
            for ( int j = i; j <= maxn; j += i )
            {
                cnt[i] += mark[j];
            }
        }
        cnt[1] = 2;
        printf("Case #%d: %d
", _case, solve(maxn));
    }
    return 0;
}

复杂度：nlogn

类似于筛法的思想进行优化，比解法1快了不少。