bnu 34895 Elegant String(矩阵快速幂)

题目链接：bnu 34895 Elegant String

题目大意：给定n和k，表示有一个长度为n的序列，序列中的元素由0~k组成，问说有多少个串满足不包含0~k的全排列。

解题思路：矩阵快速幂，根据dp[i][j]表示说第i为有j个相同，写出递推式，根据递推式求出矩阵。

#include <cstdio>
#include <cstring>

typedef long long ll;
const ll MOD = 20140518;
const int N = 20;

struct mat {
    int r, c;
    ll s[N][N];

    void init (int k, int sign) {

        memset(s, 0, sizeof(s));
        if (sign) {
            r = c = k;
            for (int i = 1; i <= r; i++) {
                for (int j = i; j <= r; j++)
                    s[i][j] = 1;
                if (i == r)
                    continue;
                s[i+1][i] = k+1-i;
            }
        } else {
            r = k;
            c = 1;
            s[1][1] = k+1;
        }
    }

    void put() {
        for (int i = 1; i <= r; i++) {
            for (int j = 1; j <= c; j++)
                printf("%lld ", s[i][j]);
            printf("\n");
        }
    }
};

mat mul(mat a, mat b) {
    mat ans;
    memset(ans.s, 0, sizeof(ans.s));

    for (int i = 1; i <= a.r; i++) {
        for (int j = 1; j <= b.c; j++) {

            for (int x = 1; x <= a.c; x++)
                ans.s[i][j] = (ans.s[i][j] + a.s[i][x] * b.s[x][j])%MOD;
        }
    }
    ans.r = a.r;
    ans.c = b.c;
    return ans;
}

ll n;
int k;

ll solve (ll x) {
    if (x == 1)
        return k+1;
    x--;
    mat ans, cur;
    ans.init(k, 0);
    cur.init(k, 1);

    while (x) {
        if (x&1)
            ans = mul(cur, ans);

        cur = mul(cur, cur);
        x /= 2;
    }

    int sum = 0;
    for (int i = 1; i <= k; i++)
        sum = (sum + ans.s[i][1]) % MOD;
    return sum;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        scanf("%lld%d", &n, &k);
        printf("Case #%d: %lld\n", i, solve(n));
    }
    return 0;
}

bnu 34895 Elegant String(矩阵快速幂),布布扣,bubuko.com

bnu 34895 Elegant String(矩阵快速幂)

原文：http://blog.csdn.net/keshuai19940722/article/details/34505379