【BZOJ3612】[Heoi2014]平衡【計數DP】【整數拆分】

【題目鏈接】

【tunix的題解】

枚舉力矩和、拿走橡皮的個數，然後就變成整數拆分問題了。

/* Telekinetic Forest Guard */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 10005, maxk = 12;

int n, k, p, dp[maxn * maxk][maxk];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

int main() {
	for(int T = iread(); T; T--) {
		n = iread(); k = iread(); p = iread();
		if(k == 1) {
			printf("1\n");
			continue;
		}

		int w = n * (k - 1);
		dp[0][0] = 1;
		for(int i = 1; i <= w; i++) for(int j = 1; j < k; j++) {
			if(i >= j) dp[i][j] = (dp[i - j][j] + dp[i - j][j - 1]) % p;
			else dp[i][j] = 0;
			if(i > n) dp[i][j] = (dp[i][j] - dp[i - (n + 1)][j - 1] + p) % p;
		}

		int ans = 0;
		for(int i = 1; i <= w; i++) for(int j = 1; j < k; j++)
			ans = (ans + dp[i][j] * dp[i][k - j] + dp[i][j] * dp[i][k - j - 1]) % p;
		printf("%d\n", ans);
	}
	return 0;
}