AcWing 1084. 数字游戏 II

题目传送门

#include <bits/stdc++.h>

using namespace std;

const int N = 15;
const int M = 110; //mod N=0 N=100

int l, r, n;
int a[N];
int dp[N][M];

int dfs(int pos, int sum, bool limit) {
    if (pos == -1) return !(sum % n);
    if (!limit && ~dp[pos][sum]) return dp[pos][sum];
    int res = 0, up = limit ? a[pos] : 9;
    for (int i = 0; i <= up; i++)
        res += dfs(pos - 1, sum + i, limit && i == a[pos]);
    return limit ? res : dp[pos][sum] = res;
}

int calc(int x) {
    memset(dp, -1, sizeof dp);
    int pos = 0;
    while (x) a[pos++] = x % 10, x /= 10;
    return dfs(pos - 1, 0, true);
}

int main() {
    while (cin >> l >> r >> n)
        cout << calc(r) - calc(l - 1) << endl;
    return 0;
}

标签：1084,return,int,sum,pos,II,limit,dp,AcWing
来源： https://www.cnblogs.com/littlehb/p/15800397.html