HDU-4389 X mod f(x) && 2018上海大都会邀请赛J 数位dp

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 int a[10], dp[10][90][90][90];
 7 
 8 int dfs(int pos, int sum, int mod, int res, bool lim){
 9     if (pos == -1) return (sum == 0 && res == mod);
10     if (!lim && dp[pos][sum][mod][res] != -1) return dp[pos][sum][mod][res];
11     int r = lim ? a[pos] : 9;
12     int ans = 0;
13     for (int i = 0; i <= r; i++){
14         ans += dfs(pos-1, (sum*10+i)%mod, mod, res+i, lim && i == a[pos]);
15     }
16     if (!lim) dp[pos][sum][mod][res] = ans;
17     return ans;
18 }
19 
20 int solve(int x){
21     int pos = 0;
22     while (x){
23         a[pos++] = x%10;
24         x /= 10;
25     }
26     int ans = 0;
27     for (int i = 1; i <= 81; i++)
28         ans += dfs(pos-1, 0, i, 0, 1);
29     return ans;
30 }
31 
32 int main(){
33     int t, a, b, kase = 0;
34     scanf("%d", &t);
35     memset(dp, -1, sizeof dp);
36     while (t--){
37         scanf("%d%d", &a, &b);
38         printf("Case %d: %d\n", ++kase, solve(b)-solve(a-1));
39     }
40     return 0;
41 }