[luoguP1373] 小a和uim之大逃离（DP）

传送门

题解

代码

#include <cstdio>

#include <iostream>

#define N 802

#define mod 1000000007

int n, m, p, ans;

int a[N][N], f[N][N][16][2];

inline int read()

{

	int x = 0, f = 1;

	char ch = getchar();

	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;

	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';

	return x * f;

}

int main()

{

	int i, j, k;

	n = read();

	m = read();

	p = read() + 1;

	for(i = 1; i <= n; i++)

		for(j = 1; j <= m; j++)

			f[i][j][(a[i][j] = read()) % p][0] = 1;

	for(i = 1; i <= n; i++)

		for(j = 1; j <= m; j++)

		{

			for(k = 0; k < p; k++)

			{

				f[i + 1][j][(k + a[i + 1][j]) % p][0] = (f[i + 1][j][(k + a[i + 1][j]) % p][0] + f[i][j][k][1]) % mod;

				f[i + 1][j][(k - a[i + 1][j] + p) % p][1] = (f[i + 1][j][(k - a[i + 1][j] + p) % p][1] + f[i][j][k][0]) % mod;

				f[i][j + 1][(k + a[i][j + 1]) % p][0] = (f[i][j + 1][(k + a[i][j + 1]) % p][0] + f[i][j][k][1]) % mod;

				f[i][j + 1][(k - a[i][j + 1] + p) % p][1] = (f[i][j + 1][(k - a[i][j + 1] + p) % p][1] + f[i][j][k][0]) % mod;

			}

			ans = (ans + f[i][j][0][1]) % mod;

		}

	printf("%d\n", ans);

	return 0;

}