hdu4722 Good Numbers（数位dp）

Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3529 Accepted Submission(s): 1128

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

   
   
   
   

    
    
    
    2
1 10
1 20
   
   
   
   

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 

题意：求a到b（并不包括b）之间共有多少个数能被10整除。

分析：基础题，dp[ i ][ j ]表示长度为 i 的数对10取模的值为 j 。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

ll s[20];
ll dp[20][10];//dp[i][j]表示长度为i的数对10取模为j

ll slove(ll x)
{
    ll t=0,sum=0;
    while (x)
    {
        s[++t]=x%10;
        x/=10;
    }
    ll ans=0,m=0;
    CL(dp);
    for (int i=t; i>0; i--)//最高位开始枚举
    {
        for (int j=0; j<10; j++)//没有界限，枚举所有
            for (int k=0; k<10; k++)
            dp[i][(j+k)%10]+=dp[i+1][j];
        for (int j=0; j<s[i]; j++)//有界限，如上一位为1，该位为2；而上一位已经是1了，所以该位只能取到2
            dp[i][(j+m)%10]++;
        m = (m+s[i])%10;//保存余数
    }
    if (!m) dp[1][0]++;
    return dp[1][0];
}

int main ()
{
    int T,ii=1;
    ll a,b;
    scanf ("%d",&T);
    while (T--)
    {
        scanf ("%lld%lld",&a,&b);
        printf ("Case #%d: %lld\n",ii++,slove(b)-slove(a-1));
    }
    return 0;
}