[数组hash] PKU 3274 Gold Balanced Lineup

题解；

数组hash：hash的是排列，不能用相加、相乘取模，常用hash函数的资料；

# include <stdio.h>
# include <string.h>

# define get(x, i) ((((x)>>(i))&0x1) ? 1:0)

# define N 100005
# define K 35
# define MOD 100007

int sum[N][K], c[N][K], a[N], h[MOD << 3];

int Max(int x, int y)
{
    return x > y ? x : y;
}

int eql(int *s, int *t, int k)
{
    int i;
    for (i = 0; i < k; ++i)
        if (s[i] != t[i]) return 0;
    return 1;
}

int hash(int *s, int k)
{
    int i;
    unsigned int h = 0, g;
    for (i = 0; i < k; ++i)
    {
        h = (h<<4)+s[i];
        g = h&0xF0000000;
        if (g) h^=g>>24;
        h &= ~g;
    }
    return (h%MOD+MOD)%MOD;
}

int main()
{
    int n, k, i, j, ans = 0, t, x;
    scanf("%d%d", &n, &k);
    memset(sum[0], 0, sizeof(sum[0]));
    memset(h, -1, sizeof(h));
    h[0] = 0;
    for (i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
        for (j = 0; j < k; ++j)
        {
            sum[i][j] += (sum[i-1][j]+get(a[i], j));
            c[i][j] = sum[i][j] - sum[i][0];
        }
        t = hash(c[i], k);
        while (h[t] != -1)
        {
            x = h[t];
            for (j = 0; j < k; ++j)
                if (c[i][j] != c[x][j]) break;
            if (j >= k)    {ans = Max(ans, i-x);break;}
            ++t;
        }
        if (h[t] == -1) h[t] = i;
    }
    printf("%d\n", ans);
    
    return 0;
}