bzoj 4321 queue2 dp
阿新 • • 發佈:2018-11-11
簡單$dp$,考慮從大到小插入
設個狀態$f[i][j][0 / 1]$表示$i$和$i - 1$是否相鄰即可
轉移看程式碼就能看懂吧...
複雜度$O(n^2)$
#include <set> #include <vector> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> namespace remoon { #define ri register int #definerep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++) #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --) #define gc getchar inline int read() { int p = 0, w = 1; char c = gc(); while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); } while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc(); return p * w; } int wr[50], rw; #define pc(iw) putchar(iw) tpr inline void write(ra o, char c = '\n') { if(!o) pc('0'); if(o < 0) o = -o, pc('-'); while(o) wr[++ rw] = o % 10, o /= 10; while(rw) pc(wr[rw --] + '0'); pc(c); } } using namespace std; using namespace remoon; namespace mod_mod { #define mod 7777777 inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; } inline void dec(int &a, int b) { a -= b; if(a < 0) a += mod; } inline int Inc(int a, int b) { return (a + b >= mod) ? a + b - mod : a + b; } inline int Dec(int a, int b) { return (a - b < 0) ? a - b + mod : a - b; } inline int mul(int a, int b) { return 1ll * a * b % mod; } } using namespace mod_mod; #define sid 1005 int n; int f[sid][sid][2]; int main() { n = read(); f[1][0][0] = 1; rep(i, 0, n) rep(j, 0, i - 1) { inc(f[i + 1][j][0], mul(f[i][j][0], i - j - 1)); if(j) inc(f[i + 1][j - 1][0], mul(f[i][j][0], j)); inc(f[i + 1][j + 1][1], mul(f[i][j][0], 2)); inc(f[i + 1][j][1], f[i][j][1]); inc(f[i + 1][j + 1][1], f[i][j][1]); inc(f[i + 1][j][0], mul(f[i][j][1], i - j)); if(j) inc(f[i + 1][j - 1][0], mul(f[i][j][1], j - 1)); } write(f[n][0][0]); return 0; }
打表可$O(1)$