hdu 4722（动态规划-数位dp）

问题描述:
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 20Sample Output

Case #1: 0
Case #2: 1

题目题意:问区间[l,r]内有多少数满足数位之和能被10整除
题目分析:数位dp，dp[len][sum]，len表示位数长度，sum表示前面位数的和
分享一个博主的数位dp总结点击打开链接
代码如下:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;

ll dp[25][2000];
int num[25];

ll dfs(int len,ll sum,bool limit)
{
if (len<1) return (ll) sum%10==0;
if (!limit&&dp[len][sum]) return dp[len][sum];
int Max=limit?num[len]:9;
ll ans=0;
for (int i=0;i<=Max;i++) {
ans+=dfs(len-1,sum+i,limit&&num[len]==i) ;
}
if (!limit) dp[len][sum]=ans;
return ans;
}
ll solve(ll n)
{
int len=0;
if (n<0) return 0;
while (n) {
num[++len]=n%10;
n=n/10;
}
return dfs(len,0,true);
}
int main()
{
int t;
scanf("%d",&t);
memset (dp,0,sizeof (dp));
for (int i=1;i<=t;i++) {
ll left,right;
scanf("%lld%lld",&left,&right);
printf("Case #%d: %lld\n",i,solve(right)-solve(left-1));
}
return 0;
}