[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；
1 # include <stdio.h> 2 # include <string.h> 3 4 # define get(x, i) ((((x)>>(i))&0x1) ? 1:0) 5 6 # define N 100005 7 # define K 35 8 # define MOD 100007 9 10 int sum[N][K], c[N][K], a[N], h[MOD << 3]; 11 12 int Max(int x, int y) 13 { 14 return x > y ? x : y; 15 } 16 17 int eql(int *s, int *t, int k) 18 { 19 int i; 20 for (i = 0; i < k; ++i) 21 if (s[i] != t[i]) return 0; 22 return 1; 23 } 24 25 int hash(int *s, int k) 26 { 27 int i; 28 unsigned int h = 0, g; 29 for (i = 0; i < k; ++i) 30 { 31 h = (h<<4)+s[i]; 32 g = h&0xF0000000; 33 if (g) h^=g>>24; 34 h &= ~g; 35 } 36 return (h%MOD+MOD)%MOD; 37 } 38 39 int main() 40 { 41 int n, k, i, j, ans = 0, t, x; 42 scanf("%d%d", &n, &k); 43 memset(sum[0], 0, sizeof(sum[0])); 44 memset(h, -1, sizeof(h)); 45 h[0] = 0; 46 for (i = 1; i <= n; ++i) 47 { 48 scanf("%d", &a[i]); 49 for (j = 0; j < k; ++j) 50 { 51 sum[i][j] += (sum[i-1][j]+get(a[i], j)); 52 c[i][j] = sum[i][j] - sum[i][0]; 53 } 54 t = hash(c[i], k); 55 while (h[t] != -1) 56 { 57 x = h[t]; 58 for (j = 0; j < k; ++j) 59 if (c[i][j] != c[x][j]) break; 60 if (j >= k) {ans = Max(ans, i-x);break;} 61 ++t; 62 } 63 if (h[t] == -1) h[t] = i; 64 } 65 printf("%d\n", ans); 66 67 return 0; 68 }
