BNU 34895 elegant string (matrix fast power)

Link: BNU 34895 elegant string

Given N and K, there is a sequence with a length of n. The elements in the sequence are from 0 ~ K composition, ask how many strings meet does not contain 0 ~ K.

Solution: the matrix's fast power. According to DP [I] [J], I indicates that J is the same, write the recursive formula, and find the matrix based on the recursive formula.

#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;}