uva 11651 - Krypton Number System(矩阵快速幂)

题目链接：uva 11651 - Krypton Number System

题目大意：给定进制base，和分数score，求在base进制下，有多少个数的值为score，要求不能有连续相同的数字以及前导0.计算一个数的值即为相邻两位数的平方差和。

解题思路：因为score很大，所以直接dp肯定超时，但是即使对于base=6的情况，每次新添一个数score最大增加25（0-5），所以用dp[i][j]预处理出base平方以内的总数，然后用矩阵快速幂计算。

#include <cstdio>
#include <cstring>
#include <algorithm>

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;
}