数位dpEnigma

Posted 2021-01-03 stolf

http://codeforces.com/group/aUVPeyEnI2/contest/227850
E

与一般数位dp不同，保存的是能否满足条件，而非记录方案数

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
typedef long long ll;
int len, n, ans[1005], dp[1005][1005];
char s[1005];
bool found = false;
bool dfs(int pos, int j) {
    if (pos == len) {
        if (j == 0) {
            for (int i = 0; i < len; i++) {
                printf("%d", ans[i]);
            }
            puts("");
            exit(0);
        }
        return false;
    }
    if (dp[pos][j] != -1) {
        return dp[pos][j];
    }
    if (s[pos] != ‘?‘) {
        ans[pos] = s[pos] - ‘0‘;
        dp[pos][j] = dfs(pos + 1, (j * 10 + ans[pos]) % n);
    }
    else for (int i = pos ? 0 : 1; i <= 9; i++) {
        ans[pos] = i;
        dp[pos][j] = dfs(pos + 1, (j * 10 + i) % n);
    }
    if (dp[pos][j] == -1) {
        dp[pos][j] = 0;
    }
    return dp[pos][j];
}
int main() {
    scanf("%s %d", s, &n);
    len = strlen(s);
    memset(dp, -1, sizeof dp);
    if (!dfs(0, 0)) {
        puts("*");
    }
}