hdu 5456 Matches Puzzle Game(记忆化搜索)

题目链接:hdu 5456 Matches Puzzle Game

解题思路

式子可以变换成A=B+C，从低位处理到高位， dp[i][j][b][c] 表示到第i位，j有没进位，b为数字B是否已经到达最高为，c为数字C是否已经到达最高位。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int need[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int maxn = 505;

int N, M;
ll dp[maxn][2][2][2];

ll dfs (int n, int p, int b, int c) {
    if (n < 0) return 0;

    ll& ret = dp[n][p][b][c];

    if (ret != -1) return ret;
    if (((n == need[1] && p) || (n == 0 && p == 0)) && b && c)
        return ret = 1;
    if (n == 0) return 0;
    ret = 0;

    if (b == 0 && c == 0) {
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < 10; j++) {
                int x = (i + j + p) / 10, y = (i + j + p) % 10;
                ret += dfs(n-need[i]-need[j]-need[y], x, 0, 0);
                if (i)
                    ret += dfs(n-need[i]-need[j]-need[y], x, 1, 0);
                if (j)
                    ret += dfs(n-need[i]-need[j]-need[y], x, 0, 1);
                if (i&&j)
                    ret += dfs(n-need[i]-need[j]-need[y], x, 1, 1);
                ret %= M;
            }
        }
    } else if (b == 0) {
        for (int i = 0; i < 10; i++) {
            int x = (i + p) / 10, y = (i + p) % 10;
            ret += dfs(n-need[i]-need[y], x, 0, 1);
            if (i)
                ret += dfs(n-need[i]-need[y], x, 1, 1);
            ret %= M;
        }
    } else if (c == 0) {
        for (int i = 0; i < 10; i++) {
            int x = (i + p) / 10, y = (i + p) % 10;
            ret += dfs(n-need[i]-need[y], x, 1, 0);
            if (i)
                ret += dfs(n-need[i]-need[y], x, 1, 1);
            ret %= M;
        }
    }
    return ret % M;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        scanf("%d%d", &N, &M);
        memset(dp, -1, sizeof(dp));
        printf("Case #%d: %lld\n", kcas, dfs(N-3, 0, 0, 0));
    }
    return 0;
}