Ultraviolet A 11651-kryton Number System (DP + matrix power)

Ultraviolet A 11651-kryton Number System

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Question: Give a base in hexadecimal notation. If a score is used to calculate the base, how many of them meet the following conditions:
1. No consecutive numbers
2. No leading zero
3. The score is score, and the score is calculated by the sum of the square differences of adjacent numbers.

Idea: Start with DP. DP [I] [J] indicates the number of I, and the last number is the number of J. Then, the status is transferred and the status that can be formed by the previous step is introduced, that is, to DP [(B-1) * (B-1)] [X].

Then we can find that all the subsequent states can be uniformly transferred from the previous states, so that we can use the matrix's rapid power for optimization. the time complexity is O (n ^ 3log (score ))

In this case, the matrix will time out when the power posture is not beautiful, and needs to be optimized when the matrix is multiplied.

Code:

#include <cstdio>#include <cstring>typedef unsigned long long ll;const int N = 155;const ll MOD = (1ULL<<32);int n;struct Mat {    ll v[N][N];    Mat() {memset(v, 0, sizeof(v));    }    void init() {memset(v, 0, sizeof(v));    }    void init_one() {memset(v, 0, sizeof(v));for (int i = 0; i < n; i++)    v[i][i] = 1;    }    Mat operator * (Mat c) {Mat ans;for (int k = 0; k < n; k++) {    for (int i = 0; i < n; i++) {if (!v[i][k]) continue;for (int j = 0; j < n; j++)    ans.v[i][j] = (ans.v[i][j] + v[i][k] * c.v[k][j] % MOD) % MOD;    }}return ans;    }} B;Mat pow_mod(Mat x, int k) {    Mat ans;    ans.init_one();    while (k) {if (k&1) ans = ans * x;x = x * x;k >>= 1;    }    return ans;}ll dp[40][10], A[N];int t, b, s, m;void init() {    scanf("%d%d", &b, &s);    m = (b - 1) * (b - 1);    n = m  * b;    memset(dp, 0, sizeof(dp));    for (int i = 0; i < b; i++)dp[0][i] = 1;    for (int i = 1; i <= m; i++) {for (int j = 0; j < b; j++) {    for (int k = 0; k < b; k++) {if (k == j) continue;int tmp = (k - j) * (k - j);if (i - tmp < 0) continue;dp[i][j] = (dp[i][j] + dp[i - tmp][k]) % MOD;    }}    }}void build() {    B.init();    for (int i = b; i < n; i++)B.v[i][i - b] = 1;    for (int i = 1; i <= m; i++) {for (int j = 0; j < b; j++) {    for (int k = 0; k < b; k++) {if (j == k) continue;if (i + (k - j) * (k - j) == m + 1) {    B.v[(i - 1) * b + j][n - b + k] = 1;}    }}    }}ll solve() {    ll ans = 0;    if (s <= m) {for (int i = 1; i < b; i++)    ans = (ans + dp[s][i]) % MOD;return ans;    }    for (int i = 1; i <= m; i++) {for (int j = 0; j < b; j++)    A[(i - 1) * b + j] = dp[i][j];    }    build();    B = pow_mod(B, s - m);    for (int i = n - b + 1; i < n; i++) {for (int j = 0; j < n; j++)    ans = (ans + A[j] * B.v[j][i] % MOD) % MOD;    }    return ans;}int main() {    int cas = 0;    scanf("%d", &t);    while (t--) {init();printf("Case %d: %llu\n", ++cas, solve());    }    return 0;}