HDU 4722Good Numbers（数位dp）

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本文介绍了一种算法，用于计算指定范围内所有好数字的数量。所谓好数字是指其各位数字之和能被10整除的数字。通过递归深度优先搜索的方法，文章详细解释了如何高效地解决这一问题，并提供了完整的C++实现代码。

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Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5241 Accepted Submission(s): 1658

Problem Description

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10 ¹⁸).

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

Sample Input

Sample Output

  
   Case #1: 0
Case #2: 1

   
    
     Hint
    
The answer maybe very large, we recommend you to use long long instead of int.

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

想法：对每位当前位%10，在dfs遍历

代码：

#include<stdio.h>
#include<string.h>
typedef long long ll;
ll dp[20][105];
int a[20];
ll dfs(int pos,int sum,bool limit)
{
    if(pos==-1) return sum==0;
    if(!limit&&dp[pos][sum]!=-1) return dp[pos][sum];
    int up=limit?a[pos]:9;
    ll tmp=0;
    for(int i=0;i<=up;i++)
    {
        tmp+=dfs(pos-1,(sum+i)%10,limit&&(i==up));//
    }
    if(!limit) dp[pos][sum]=tmp;
    return tmp;
}
ll solve(ll x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,0,1);
}
int main()
{
    int T,ws=1;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while(T--)
    {
      ll n,m;
      scanf("%lld %lld",&n,&m);
      printf("Case #%d: %lld\n",ws++,solve(m)-solve(n-1));
    }
    return 0;
}