hdu 4722 Good Numbers( 数位dp入门)

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3467    Accepted Submission(s): 1099

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.

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

题意：给定区间A到B（用 long long），求其间好数有多少，好数是指每位数字加起来的和对10取余结果是0。

当时不明白何为学长讲的边界，另外数位dp还有递推版，但是递推太容易出错，还是记忆化搜索比较容易理解～

 #include<iostream>
 #include<vector>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <math.h>
 #include<algorithm>
 #define ll long long
 #define eps 1e-8
 using namespace std;
 ll dp[][]; // 统计第几位数余数为几有多少个
 int num[]; //存放每位数字
 ll dfs(int site,int mod,int f)
 {
     if(site == )
         return mod ==  ? :; // 递归结束条件，并判断最后一位时是否余数为0
     if(!f && dp[site][mod] != -) //记忆化搜索
         return dp[site][mod];
     int len = f == ?num[site]:; //若当前位为边界則当前位只枚举0-当前，否则枚举0-9
     ll ans = ;
     for(int i = ; i <= len; i++)
         ans += dfs(site-,(mod+i)%,f && i == len);//f判断是否为边界
     if(!f)
         dp[site][mod] = ans; //边界不能存，因为之前计算的是不看边界的，便于以后用到，如果存了，就是一个错误的统计
     return ans;
 }
 ll solve(ll digit)
 {
     memset(num,,sizeof(num));
     int l = ;
     while(digit)
     {
         num[++l] = digit%;
         digit /= ;
     }
     return dfs(l,,);
 }
 int main(void)
 {
     int t,cnt = ;
     ll a,b;
     memset(dp,-,sizeof(dp));
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lld %lld",&a,&b);
         printf("Case #%d: %lld\n",cnt++,solve(b)-solve(a-));
     }
     return ;
 }