Ultraviolet A 11651-kryton Number System (matrix power)

Ultraviolet A 11651-kryton Number System (matrix power)

Link to the Q & A 11651-kryton Number System

Given base, and score, evaluate the number of scores in base. The same consecutive number and leading 0 are not required. calculate the square difference between two adjacent digits.

Solution: Because the score is large, the direct dp definitely times out. However, even if the base is 6, a new number of scores can be added up to 25 (0-5) at a time ), therefore, we use dp [I] [j] to pre-process the total number within the base square, and then use the Matrix to calculate the power quickly.

#include 
  
   #include 
   
    #include using namespace std;typedef unsigned long long ll;const int maxn = 155;const ll MOD = 1ll<<32;struct Mat {    int r, c;    ll arr[maxn][maxn];    Mat (int r = 0, int c = 0) { set(r, c); }    void set(int r, int c) {        this->r = r;        this->c = c;        memset(arr, 0, sizeof(arr));    }    Mat operator * (const Mat& u) {        Mat ret(r, u.c);        for (int k = 0; k < c; k++) {            for (int i = 0; i < r; i++) {                if (arr[i][k] == 0)                    continue;                for (int j = 0; j < u.c; j++)                    ret.arr[i][j] = (ret.arr[i][j] + arr[i][k] * u.arr[k][j]) % MOD;            }        }        return ret;    }};int base, N;ll dp[maxn][maxn], score;void init () {    scanf("%d%llu", &base, &score);    N = (base-1) * (base-1);    memset(dp, 0, sizeof(dp));    for (int i = 0; i <= N; i++)        dp[0][i] = 1;    for (int i = 0; i < N; i++) {        for (int j = 0; j < base; j++) {            for (int k = 0; k < base; k++) {                int f = (j - k) * (j - k);                if (i + f > N || f == 0)                    continue;                dp[i+f][j] = (dp[i+f][j] + dp[i][k]) % MOD;            }        }    }}Mat change () {    Mat ret(N*base, 1);    for (int i = 1; i <= N; i++)        for (int j = 0; j < base; j++)            ret.arr[(i-1)*base+j][0] = dp[i][j];    return ret;}Mat build () {    int n = N * base;    Mat x(n, n);    for (int i = base; i < n; i++)        x.arr[i-base][i] = 1;    for (int i = 0; i < base; i++) {        for (int j = 0; j < base; j++) {            if (i == j)                continue;            int k = N - (i-j) * (i-j);            x.arr[(N-1)*base+i][k*base+j] = 1;        }    }    return x;}Mat pow_mat (Mat ret, int n) {    Mat x = build();    while (n) {        if (n&1)            ret = x * ret;        x = x * x;        n >>= 1;    }    return ret;}ll solve () {    ll ans = 0;    if (score <= N) {        for (int i = 1; i < base; i++)            ans = (ans + dp[score][i]) % MOD;        return ans;    }    Mat ret = change();    ret = pow_mat(ret, score-N);    for (int i = 1; i < base; i++)        ans = (ans + ret.arr[(N-1)*base+i][0]) % MOD;    return ans;}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        init();        printf("Case %d: %llu\n", kcas, solve());    }    return 0;}