2018 ACM 国际大学生程序设计竞赛上海大都会赛 J [Beautiful Number]-优快云博客

链接：https://www.nowcoder.com/acm/contest/163/J
来源：牛客网

时间限制：C/C++ 8秒，其他语言16秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

NIBGNAUK is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by the sum of its digits.
We will not argue with this and just count the quantity of beautiful numbers from 1 to N.

输入描述:

The first line of the input is T(1≤ T ≤ 100), which stands for the number of test cases you need to solve.
Each test case contains a line with a positive integer N (1 ≤ N ≤ $10^{12}$ ).

输出描述:

For each test case, print the case number and the quantity of beautiful numbers in [1, N].

示例1

输入

2
10
18

输出

Case 1: 10
Case 2: 12

Solution

数位DP。与BZOJ1799如出一辙，数据小了些，测试样例变多组而已。
考虑枚举1..12*9这108个可能的除数，因1.. $10^{12}$ 之内数字的各位数字之和最小是1最大是12*9=108；
还是从高位到低位枚举，下标pos-1到0，每次枚举的数位和作为令符合要求的被除数可以整除的模数mod，同时也是各位数字之和sum.
dp[pos][sum][val][0]: 表示到了第pos位，各位数字和是sum，前面pos位形成的数模当前枚举的mod为val，且前pos位全都和原数x相等的数的个数；
dp[pos][sum][val][1]: 表示到了第pos位，各位数字和是sum，前面pos位形成的数模当前枚举的mod为val，且前pos位全都小于原数x的数的个数.

#include <bits/stdc++.h>

#define ll long long
using namespace std;
ll dp[12][110][110][2];
int bit[15];

ll dfs(int pos, int sum, int val, int mod, bool limit) {
    if (sum - 9 * pos > 9) return 0;
    if (pos == -1) return sum == 0 && val == 0;
    if (dp[pos][sum][val][limit] != -1) return dp[pos][sum][val][limit];
    int up = limit ? bit[pos] : 9;
    ll ans = 0;
    for (int i = 0; i <= up; i++) {
        if (sum < i) break;
        ans += dfs(pos - 1, sum - i, (val * 10 + i) % mod, mod, limit && i == bit[pos]);
    }
    return dp[pos][sum][val][limit] = ans;
}

ll solve(ll n) {
    int l = 0;
    while (n) {
        bit[l++] = n % 10;
        n /= 10;
    }
    ll res = 0;
    for (int i = 1; i <= l * 9; ++i) {
        memset(dp, -1, sizeof(dp));
        res += dfs(l - 1, i, 0, i, 1);
    }
    return res;
}

int T;
ll x;

int main() {
    //freopen("../in", "r", stdin);
    scanf("%d", &T);
    for (int I = 1; I <= T; ++I) {
        printf("Case %d: ",I);
        scanf("%lld", &x);
        if (x == 1000000000000) printf("%lld\n", solve(999999999999) + 1);
        else printf("%lld\n", solve(x));
    }
    return 0;
}