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May 5, 2017 03:56
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BZOJ1494 failing dp
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#include <cstdio> | |
#include <algorithm> | |
using namespace std; | |
typedef long long ll; | |
const ll mod = 65521; | |
inline ll pow(ll a, ll b) { | |
ll ret = 1; | |
for(; b; b >>= 1, a = a * a % mod) if(b & 1) ret = ret * a % mod; | |
return ret; | |
} | |
const int N = 100 + 5; | |
int k, n; | |
ll mt[N][N]; | |
int main() { | |
scanf("%d%d", &k, &n); | |
for(int i = 1; i <= n; ++i) for(int j = i + 1; j <= min(n, i + k); ++j) { | |
--mt[i][j], --mt[j][i]; | |
++mt[i][i], ++mt[j][j]; | |
} | |
ll ans = 1; | |
for(int i = 1; i < n; ++i) { | |
int jid = -1; | |
for(int j = i; j < n; ++j) if(mt[j][i]) { jid = j; break; } | |
if(jid != i) for(int j = 1; j < n; ++j) swap(mt[i][j], mt[jid][j]); | |
ans = ans * mt[i][i] % mod; | |
ll i_mtii = pow(mt[i][i], mod - 2); | |
for(int j = 1; j < n; ++j) mt[i][j] = mt[i][j] * i_mtii % mod; | |
for(int kk = 1; kk < n; ++kk) if(kk != i) { | |
ll mtkki = mt[kk][i]; | |
for(int j = 1; j < n; ++j) mt[kk][j] = (mt[kk][j] - mt[i][j] * mtkki % mod + mod) % mod; | |
} | |
} | |
printf("%lld\n", (ans % mod + mod) % mod); | |
return 0; | |
} |
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#include <cstdio> | |
#include <algorithm> | |
using namespace std; | |
typedef long long ll; | |
const ll mod = 65521; | |
inline ll pow(ll a, ll b) { | |
ll ret = 1; | |
for(; b; b >>= 1, a = a * a % mod) if(b & 1) ret = ret * a % mod; | |
return ret; | |
} | |
const int N = 100 + 5, K = 5, S = (1 << (2 * K)) + 1; | |
int kbd, n; | |
ll f[N][S]; | |
inline int g_mt(int i, int j) { | |
if(i == j) return min(n, i + kbd) - max(1, i - kbd); | |
else return abs(i - j) <= kbd ? mod - 1 : 0; | |
} | |
int popcount[S]; | |
int main() { | |
scanf("%d%d", &kbd, &n); | |
f[1][(1 << kbd) - 1] = 1; | |
for(int s = 1; s < (1 << (kbd * 2)); ++s) popcount[s] = popcount[s >> 1] + (s & 1); | |
for(int i = 1; i < n - 1; ++i) for(int s = 0; s < (1 << (kbd * 2)); ++s) if(f[i][s]) { | |
for(int j = -kbd; j <= kbd; ++j) if(i + j >= 1 && i + j < n && !(s >> (j + kbd) & 1) && ((s | 1 << (j + kbd)) & 1)) { | |
ll &nxt = f[i + 1][(s | 1 << (j + kbd)) >> 1], ths = f[i][s]; | |
nxt = (nxt + ths * g_mt(i, i + j) % mod * pow(mod - 1, popcount[s >> (j + kbd + 1)])) % mod; | |
} | |
} | |
printf("%lld\n", f[n - 1][(1 << kbd) - 1]); | |
return 0; | |
} |
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