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 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
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


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