uva11361数位dp

挺裸的 ，只要注意到当k超过9*10  就直接输出0就可以了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include <math.h>
using namespace std;typedef long long LL;
LL dp[15][150][150];
LL k;
LL path[1000];LL gao(LL x, LL sum, LL mod, LL flag)
{if (~dp[x][sum][mod]&&!flag) return  dp[x][sum][mod];if (x == 0){if (sum%k == 0 && mod == 0) return 1;else return 0;}LL  bound = flag ? path[x] : 9;LL ans = 0;for (LL i = 0; i <= bound; i++){// if((mod* 10+ i ) % k > 100) continue;ans += gao(x - 1, sum + i, (mod * 10 + i) % k, flag && (i == bound));}return flag?ans : dp[x][sum][mod] = ans;
}LL solve(LL x)
{LL ret = 0;while (x){path[++ret] = x % 10;x /= 10;}return gao(ret, 0, 0, 1);
}int main()
{LL a, b;LL Icase;cin >> Icase;while (Icase--){cin >> a >> b >> k;if(k>90){cout<<0<<endl;continue;}memset(dp,-1,sizeof(dp));cout << solve(b) - solve(a - 1) << endl;}return 0;
}