Hard
Number of Distinct Roll Sequences — C++
Full explanation · Time O(6^3 * n) · Space O(6^2)
// Time: O(6^3 * n)
// Space: O(6^2)
// dp
class Solution {
public:
int distinctSequences(int n) {
static const int MOD = 1e9 + 7;
if (n == 1) {
return 6;
}
vector<vector<int>> dp(6, vector<int>(6));
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
if (i != j && gcd(i + 1, j + 1) == 1) {
dp[i][j] = 1;
}
}
}
for (int _ = 0; _ < n - 2; ++_) {
vector<vector<int>> new_dp(6, vector<int>(6));
for (int i = 0; i < 6; ++i) {
for (int j = 0; j < 6; ++j) {
if (!dp[i][j]) {
continue;
}
for (int k = 0; k < 6; ++k) {
if (!dp[j][k]) {
continue;
}
if (k != i) {
new_dp[i][j] = (new_dp[i][j] + dp[j][k]) % MOD;
}
}
}
}
dp = move(new_dp);
}
return accumulate(cbegin(dp), cend(dp), 0,
[&](int total, const auto& x) {
return (total + accumulate(cbegin(x), cend(x), 0,
[&](int total, int x) {
return (total + x) % MOD;
})) % MOD;
});
}
};