HDU 4722 Good Numbers（数位DP）

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=4722

Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4395 Accepted Submission(s): 1427

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 <= 1018).

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.

【中文题意】问你一个范围内有多少好数，好数的定义如下：他的所有位数之和能被10整除。
【思路分析】非常典型的数位DP，记录这些位数之和就行了。
【AC代码】

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define LL long longLL dp[20][12],dight[20];LL dfs(LL pos,LL pre,LL limit){    if(pos==-1)//终止条件    {        return !(pre%10);//终止时判断是否是10的倍数    }    if(!limit&&dp[pos][pre]!=-1)//如果之前计算过，现在就不需要计算了    {        return dp[pos][pre];    }    LL re=0;    LL tmp=limit?dight[pos]:9;    for(LL i=0;i<=tmp;i++)    {            re+=dfs(pos-1,(pre+i)%10,limit&&i==tmp);    }    if(!limit)    {        dp[pos][pre]=re;    }    return re;}LL cal(LL n){    LL len=0;    while(n)    {        dight[len++]=n%10;        n/=10;    }    return dfs(len-1,0,1);}int main(){    memset(dp,-1,sizeof(dp));    int t,iCase=0;    scanf("%d",&t);    while(t--)    {        LL a,b;        scanf("%lld%lld",&a,&b);        printf("Case #%d: %lld\n",++iCase,cal(b)-cal(a-1));    }    return 0;}