HDU 5456 Matches Puzzle Game——数位dp

先把式子转化为B+C=A；然后用dp[i][j][0/1][0/1][0/1]表示A用了i根火柴棒，从低到高递推到第j位，上一位是否有进位，B是否用到了最高位，C是否用到了最高位的情况总数，最后统计一下dp[n][枚举位数][0][1][1]的和就好了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int num[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
int T, n, mod;
ll dp[510][260][2][2][2];
int main() {
    scanf("%d", &T);
    for (int kase = 1; kase <= T; kase++) {
        scanf("%d%d", &n, &mod);
        n -= 3;
        int bit = n / 2;
        memset(dp, 0, sizeof(dp));
        dp[0][0][0][0][0] = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < bit; j++) {
                for (int k = 0; k < 2; k++) {
                    if (dp[i][j][k][0][0] != 0) {
                        for (int x = 0; x <= 9; x++) {
                            for (int y = 0; y <= 9; y++) {
                                int t1 = num[(x+y+k)%10] + num[x] + num[y];
                                int t2 = (x+y+k)/10;
                                if (i + t1 > n) continue;
                                dp[i+t1][j+1][t2][0][0] = (dp[i+t1][j+1][t2][0][0]+dp[i][j][k][0][0])%mod;
                                if (y != 0) dp[i+t1][j+1][t2][0][1] = (dp[i+t1][j+1][t2][0][1]+dp[i][j][k][0][0])%mod;
                                if (x != 0) dp[i+t1][j+1][t2][1][0] = (dp[i+t1][j+1][t2][1][0]+dp[i][j][k][0][0])%mod;
                                if (x != 0 && y != 0) dp[i+t1][j+1][t2][1][1] = (dp[i+t1][j+1][t2][1][1]+dp[i][j][k][0][0])%mod;
                            }
                        }
                    }
                    if (dp[i][j][k][0][1] != 0) {
                        for (int x = 0; x <= 9; x++) {
                            int t1 = num[(x+k)%10] + num[x];
                            int t2 = (x+k)/10;
                            if (i + t1 > n) continue;
                            dp[i+t1][j+1][t2][0][1] = (dp[i+t1][j+1][t2][0][1]+dp[i][j][k][0][1])%mod;
                            if (x != 0) dp[i+t1][j+1][t2][1][1] = (dp[i+t1][j+1][t2][1][1]+dp[i][j][k][0][1])%mod;
                        }
                    }
                    if (dp[i][j][k][1][0] != 0) {
                        for (int y = 0; y <= 9; y++) {
                            int t1 = num[(y+k)%10] + num[y];
                            int t2 = (y+k)/10;
                            if (i + t1 > n) continue;
                            dp[i+t1][j+1][t2][1][0] = (dp[i+t1][j+1][t2][1][0]+dp[i][j][k][1][0])%mod;
                            if (y != 0) dp[i+t1][j+1][t2][1][1] = (dp[i+t1][j+1][t2][1][1]+dp[i][j][k][1][0])%mod;
                        }
                    }
                    if (dp[i][j][k][1][1] != 0) {
                        if (k == 1 && i + num[1] <= n) dp[i+num[1]][j+1][0][1][1] = (dp[i+num[1]][j+1][0][1][1]+dp[i][j][k][1][1])%mod;
                    }
                }
            }
        }
        ll ans = 0;
        for (int i = 0; i <= bit; i++) {
            ans = (ans + dp[n][i][0][1][1]) % mod;
        }
        printf("Case #%d: %d\n", kase, ans);
    }
    return 0;
}