题链：http://acm.hdu.edu.cn/showproblem.php?pid=4389

X mod f(x)

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Here is a function f(x):
　　　int f ( int x ) {
　　　    if ( x == 0 ) return 0;
　　　    return f ( x / 10 ) + x % 10;
　　　}

　　　Now, you want to know, in a given interval [A, B] (1 <= A <= B <= 10⁹), how many integer x that mod f(x) equal to 0.

Input

　　　The first line has an integer T (1 <= T <= 50), indicate the number of test cases.

　　　Each test case has two integers A, B.

Output

　　　For each test case, output only one line containing the case number and an integer indicated the number of x.

Sample Input

2
1 10
11 20

Sample Output

Case 1: 10
Case 2: 3

数位dp学习链接：http://blog.csdn.net/cyendra/article/details/38087573

题意：相当于问区间内有多少数满足 X%（∑xi）==0。∑xi 是数字X的数位和。

做法：由于最多9位数，所以能够枚举∑xi，最大为81。 然后就是数位dp了。

sum是数位和，nwmod是取模结果，mod 是枚举的模

当数位和sum==mod并且，nwmod最后==0，成立计数。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map> 

typedef long long LL;
const int maxn=81;
int dig[maxn];
int f[10][maxn][maxn][maxn];
//nwmod  数取模后 sum 数位和
LL dfs(int pos,int nwmod,int sum,int mod,int limit)
{
	if (pos<0) return sum==mod&&nwmod==0;
	if (!limit&&f[pos][nwmod][sum][mod]!=-1)
		return f[pos][nwmod][sum][mod];
	LL res=0;
	int last=limit?dig[pos]:9;
	for (int i=0;i<=last;i++)
	{
		res+=dfs(pos-1,(nwmod*10+i)%mod,sum+i,mod,limit&&(i==last));
	}
	if (!limit) f[pos][nwmod][sum][mod]=res;
	return res;
}

LL solve(LL n){
	int len=0;
	while (n)
	{
		dig[len++]=n%10;
		n/=10;
	}
	LL ans=0;
	for(int i=1;i<=81;i++)//枚举最后的mod
	{
		ans+=dfs(len-1,0,0,i,1);
	}
	return ans;
}
int main()
{
	int n;
	int t;
	int cas=1;
	scanf("%d",&t);
	int a,b;
	memset(f,-1,sizeof f);
	while(t--)
	{
		scanf("%d%d",&a,&b);
		if(a>b)
			swap(a,b);
		printf("Case %d: %I64d\n",cas++,solve(b)-solve(a-1));
	}
	return 0;
}

/*
2
1 10
11 20 

Sample Output
Case 1: 10
Case 2: 3
*/