洛谷P4104 [HEOI2014]平衡(dp 组合数学)

#include<bits/stdc++.h>
#define Fin(x) freopen(#x".in", "r", stdin);
using namespace std;
const int MAXN = 50001;
int mod;
template<typename A, typename B> inline bool chmax(A &x, B y) {return x < y ? x = y, 1 : 0;}
template<typename A, typename B> inline bool chmin(A &x, B y) {return x > y ? x = y, 1 : 0;}
template<typename A, typename B> inline A mul(A x, B y) {return 1ll * x * y % mod;}
template<typename A, typename B> inline void add2(A &x, B y) {x = x + y >= mod ? x + y - mod : x + y;}
template<typename A, typename B> inline int add(A &x, B y) {return x + y >= mod ? x + y - mod : x + y;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = mul(base, a);
        a = mul(a, a);  p >>= 1;
    }
    return base;
}
int inv(int x) {
    return fp(x, mod - 2);
}
int f[101][100001];
void solve() {
    int N = read(), K = read(); mod = read();
    memset(f, 0, sizeof(f));
    f[0][0] = 1;
    for(int j = 1; j <= (N + 1) * K; j++)
        for(int i = 1; i <= min(j, K); i++) {
            f[i][j] = add(f[i][j - i], f[i - 1][j - i]);
            if(j >= 2 * N + 2) add2(f[i][j], -f[i - 1][j - (2 * N + 2)] + mod);
        }
    cout << f[K][(N + 1) * K] << '\n';
}
signed main() {
    for(int T = read(); T--; solve());
    return 0;
}