luogu1373_小a和uim之大逃离 多维dp

传送门

巧妙之处在于dp的设计只用设计差值即可，因此不会mle，枚举的顺序问题也解决了

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define rep(i, a, b) for (int i = a; i <= b; ++i)
 4 
 5 const int N = 801;
 6 const long long Mo = 1000000007;
 7 
 8 int n, m, k, a[N][N], dp[N][N][17][2];
 9 
10 int main() {
11     scanf("%d%d%d", &n, &m, &k); ++k;
12     rep(i, 1, n) rep(j, 1, m) {
13         scanf("%d", &a[i][j]);
14         dp[i][j][a[i][j] % k][0] = 1;
15     }
16     
17     rep(i, 1, n) rep(j, 1, m) rep(p, 0, k) {
18         dp[i][j][p][0] = (dp[i][j][p][0] + dp[i - 1][j][(p - a[i][j] + k) % k][1]) % Mo;
19         dp[i][j][p][0] = (dp[i][j][p][0] + dp[i][j - 1][(p - a[i][j] + k) % k][1]) % Mo;
20         dp[i][j][p][1] = (dp[i][j][p][1] + dp[i - 1][j][(p + a[i][j]) % k][0]) % Mo;
21         dp[i][j][p][1] = (dp[i][j][p][1] + dp[i][j - 1][(p + a[i][j]) % k][0]) % Mo;
22     }
23 
24     long long ans = 0;
25     rep(i, 1, n) rep(j, 1, m) {
26         ans = (ans + dp[i][j][0][1]) % Mo;
27     }
28 
29     printf("%lld\n", ans);
30 
31     return 0;
32 }

 

posted @ 2018-12-09 15:17  euix  阅读(145)  评论(0)  编辑  收藏  举报