[ABC135D] Digits Parade

• 题解：又是dp想不出来的。\(dp_{i, j}\) 代表了，第 \(i\) 位时（从左到右），余数为 \(j\) 的数量。然后发现，状态转移的时候，如果 $$s_i = ?$$ 则

\[dp_{i,(j * 10 + k) \% 13} =\sum dp_{i-1, (j * 10 + k) \% 13} (k \in [0, 9]) \]

否则直接

\[dp_{i,(j * 10 + s[i] - '0') \% 13} =\sum dp_{i-1, (j * 10 +s[i] - '0') \% 13} \]

• 代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const ll N = 1e6 + 9, mod = 1e9 + 7;
int n;
char s[N];
ll dp[15][N];
ll q_mi(ll a, ll k, ll p) {
    ll ret = 1;
    ll x = a;
    while (k) {
        if (k & 1) {
            (ret *= x )%= p;
        }
        (x *= x)%=p;
        k >>= 1;
    }
    return ret;
}
int main() {
    cin >> (s + 1);
    int n = strlen(s + 1);
    dp[0][0] = 1;
    for (int i = 1; i <= n; i ++) {
        if (s[i] == '?') {
            for (int k = 0; k <= 9; k ++)
            for (int j = 0; j <= 12; j ++) {
                (dp[(j * 10 + k)%13][i] += dp[j][i-1]) %= mod;
            }
        } else {
            for (int j = 0; j <= 12; j++) {
                (dp[(j * 10 + s[i] - '0') % 13][i] += dp[j][i - 1]) %= mod;
            }
        }

        //cout << dp[5][i] << endl;
    }
    cout << dp[5][n] << endl;

}