题解 | #小红删数字#

#include <iostream>
#include <vector>
#include <numeric>

using namespace std;

const int MOD = 1000000007;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    if (n == 1) {
        vector<int> result(10, 0);
        if (a[0] >= 0 && a[0] <= 9) {
            result[a[0]] = 1;
        }
        for (int i = 0; i < 10; ++i) {
            cout << result[i] << (i == 9 ? "" : " ");
        }
        cout << endl;
        return 0;
    }

    // dp[j] 表示当前后缀运算结果为 j 的方案数
    vector<int> dp(10, 0);

    // 基本情况：处理 a[n-1]
    dp[a[n - 1] % 10] = 1;

    // 从后向前递推
    for (int i = n - 2; i >= 0; --i) {
        vector<int> next_dp(10, 0);
        int current_val = a[i] % 10;
        for (int j = 0; j < 10; ++j) {
            if (dp[j] > 0) {
                // (a[i] + j) mod 10
                int res_add = (current_val + j) % 10;
                next_dp[res_add] = (next_dp[res_add] + dp[j]) % MOD;
                // (a[i] * j) mod 10
                int res_mul = (current_val * j) % 10;
                next_dp[res_mul] = (next_dp[res_mul] + dp[j]) % MOD;
            }
        }
        dp = next_dp;
    }
    
    for (int i = 0; i < 10; ++i) {
        cout << dp[i] << (i == 9 ? "" : " ");
    }
    cout << endl;

    return 0;
}

import java.util.Scanner;

public class Main {
    private static final int MOD = 1_000_000_007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
        }

        if (n == 1) {
            int[] result = new int[10];
            if (a[0] >= 0 && a[0] <= 9) {
                result[(int)a[0]] = 1;
            }
            for (int i = 0; i < 10; i++) {
                System.out.print(result[i] + (i == 9 ? "" : " "));
            }
            System.out.println();
            return;
        }

        // dp[j] 表示当前后缀运算结果为 j 的方案数
        int[] dp = new int[10];

        // 基本情况：处理 a[n-1]
        dp[(int)(a[n - 1] % 10)] = 1;

        // 从后向前递推
        for (int i = n - 2; i >= 0; i--) {
            int[] nextDp = new int[10];
            int currentVal = (int)(a[i] % 10);
            for (int j = 0; j < 10; j++) {
                if (dp[j] > 0) {
                    // (a[i] + j) mod 10
                    int resAdd = (currentVal + j) % 10;
                    nextDp[resAdd] = (nextDp[resAdd] + dp[j]) % MOD;
                    // (a[i] * j) mod 10
                    int resMul = (currentVal * j) % 10;
                    nextDp[resMul] = (nextDp[resMul] + dp[j]) % MOD;
                }
            }
            dp = nextDp;
        }
        
        for (int i = 0; i < 10; i++) {
            System.out.print(dp[i] + (i == 9 ? "" : " "));
        }
        System.out.println();
    }
}

import sys

def main():
    MOD = 1_000_000_007
    
    try:
        n_str = sys.stdin.readline()
        if not n_str: return
        n = int(n_str)
        a = list(map(int, sys.stdin.readline().split()))
    except (IOError, ValueError):
        return

    if n == 1:
        result = [0] * 10
        if 0 <= a[0] <= 9:
            result[a[0]] = 1
        print(*result)
        return

    # dp[j] 表示当前后缀运算结果为 j 的方案数
    dp = [0] * 10

    # 基本情况：处理 a[n-1]
    dp[a[n - 1] % 10] = 1
    
    # 从后向前递推
    for i in range(n - 2, -1, -1):
        next_dp = [0] * 10
        current_val = a[i] % 10
        for j in range(10):
            if dp[j] > 0:
                # (a[i] + j) mod 10
                res_add = (current_val + j) % 10
                next_dp[res_add] = (next_dp[res_add] + dp[j]) % MOD
                # (a[i] * j) mod 10
                res_mul = (current_val * j) % 10
                next_dp[res_mul] = (next_dp[res_mul] + dp[j]) % MOD
        dp = next_dp

    print(*dp)


if __name__ == "__main__":
    main()