HDU - 5362 Just A String dp

HDU - 5362

还真的就卡常啊， T1000组， 我以为有什么nb的预处理方法呢。

每个长度的贡献期望是一样的， dp出来然后计算就好了。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a % mod;
        a = 1LL * a * a % mod; b >>= 1;
    }
    return ans;
}

int n, m;
int inv[N];
int dp[N][N];

int main() {

    inv[1] = 1;
    for(int i = 2; i < N; i++) {
        inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    }
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        dp[0][0] = 1;
        for(int i = 1; i <= n; i++) {
            for(int j = (i & 1); j <= m; j += 2) {
                dp[i][j] = 0;
                if(j > 0) {
                    add(dp[i][j], 1LL * dp[i - 1][j - 1] * (m - j + 1) % mod);
                }
                if(j < m) {
                    add(dp[i][j], 1LL * dp[i - 1][j + 1] * (j + 1) % mod);
                }
            }
        }
        int ans = 0;
        for(int i = 1; i <= n; i++) {
            int tmp = power(m, n - i);
            int p = dp[i][i & 1];
            int c = n - i + 1;
            add(ans, 1LL * p * c % mod * tmp % mod);
        }
        printf("%d\n", ans);
    }
    return 0;
}

/**/