hdu 3709 数位DP

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
int a[22];
ll dp[22][22][2222];
ll dfs(int len, int idx, int sum, bool lim) {
if(!len) return !sum;
if(sum < 0) return 0;
if(!lim && ~dp[len][idx][sum]) return dp[len][idx][sum];
ll ret = 0;
int m = lim ? a[len] : 9;
for(int i = 0; i <= m; i++)
ret += dfs(len-1, idx, sum+(len-1-idx)*i, lim&&i==m);
if(!lim) dp[len][idx][sum] = ret;
return ret;
}
ll gao(ll n) {
if(n < 0) return 0;
int len = 0;
while(n) {
a[++len] = n%10;
n /= 10;
}
ll ret = 0;
for(int i = 0; i < len; i++)
ret += dfs(len, i, 0, 1);
return ret-len+1; //去掉支点不在最低位而且全是数字0的情况
}
int main() {
int cas;
ll l, r;
memset(dp, -1, sizeof(dp));
scanf("%d", &cas);
while(cas--) {
cin >> l >> r;
cout << gao(r) - gao(l-1) << endl;;
}
return 0;
}