# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
83185 | nikolapesic2802 | Beautiful row (IZhO12_beauty) | C++14 | 1058 ms | 164828 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <stdio.h>
#include <algorithm>
using namespace std;
#define int long long
#define ll long long
int a[22];
int n;
// can these two elements i and j be neighbours?
int ok[22][22];
// return number of '1' in ternary representation
int get(int x) {
int res = 0;
while (x) {
res += (x % 3 == 1);
x /= 3;
}
return res;
}
// return number of '1' in binary representation
int get1(int x) {
int res = 0;
while (x) {
res += (x % 2 == 1);
x /= 2;
}
return res;
}
/*
dp[mask][i] = mask shows which elements we took, it avoids to take element twice
i shows which element we took last, it avoids to calc wrong permutation
*/
int dp[1 << 20][20];
main() {
int n;
scanf("%lld", &n);
for (int i = 0; i < n; i++) {
int x;
scanf("%lld", &x);
a[i] = x;
// initially we can take an element i
dp[1 << i][i]++;
}
// precalc ok[i][j]
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
ok[i][j] = (get(a[i]) == get(a[j]) || get1(a[i]) == get1(a[j]));
}
}
for (int mask = 1; mask < (1 << n); mask++) {
for (int j = 0; j < n; j++) {
if ((mask >> j) & 1 ^ 1) continue;
// we push value of dp[mask][j] to front
// if element j is not taken, skip it
// because we add to the dp[mask | (1 << k)][k] the value dp[mask][j]
// we can't add dp[mask][j] if we didn't took it
for (int k = 0; k < n; k++) {
if (((mask >> k) & 1 ^ 1) && ok[j][k]) {
// if we can take an element k after element j
// we push value front
dp[mask | (1 << k)][k] += dp[mask][j];
}
}
}
}
// answer is sum of ( dp[2^n - 1][i] ): 0 <= i <= n - 1
int ans = 0;
for (int i = 0; i < n; i++) {
ans += dp[(1 << n) - 1][i];
}
printf("%lld", ans);
}
Compilation message (stderr)
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