HDU

HDU - 5852

就是个裸的LGV定理套一下， 求下行列式的值。

#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 = 100 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)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 n, k;
int a[N], b[N];
int mat[N][N];

const int FN = (int)2e5 + 7;
int F[FN], Finv[FN], inv[FN];

int solve(int a[N][N], int n) {
    int sign = 0;
    int ans = 1;
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            int x = i, y = j;
            while(a[y][i]) {
                int t = a[x][i] / a[y][i];
                for(int k = i; k < n; k++) {
                    a[x][k] = (a[x][k] - 1LL * a[y][k] * t) % mod;
                }
                swap(x, y);
            }
            if(x != i) {
                for(int k = 0; k < n; k++) swap(a[i][k], a[x][k]);
                sign ^= 1;
            }
        }
        if(a[i][i] == 0) return 0;
        else ans = 1LL * ans * a[i][i] % mod;

    }
    if(sign != 0) ans *= -1;
    if(ans < 0)ans += mod;
    return ans;
}

int calc(int x, int y) {
    if(x < 0 || y < 0) return 0;
    return 1LL * F[x + y] * Finv[x] % mod * Finv[y] % mod;
}

void prepare() {
    F[0] = Finv[0] = inv[1] = 1;
    for(int i = 2; i < FN; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    for(int i = 1; i < FN; i++) F[i] = 1LL * F[i - 1] * i % mod;
    for(int i = 1; i < FN; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod;
}

int main() {
    prepare();
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &k);
        for(int i = 0; i < k; i++) {
            scanf("%d", &a[i]);
        }
        for(int i = 0; i < k; i++) {
            scanf("%d", &b[i]);
        }
        if(n == 1) {
            puts("1");
            continue;
        }
        for(int i = 0; i < k; i++) {
            for(int j = 0; j < k; j++) {
                mat[i][j] = calc(n - 1, b[j] - a[i]);
            }
        }
        printf("%d
", solve(mat, k));
    }
    return 0;
}

/*
*/