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[20][20]; // 统计第几位数余数为几有多少个
int num[20]; //存放每位数字
ll dfs(int site,int mod,int f)
{
    if(site == 0)
        return mod == 0 ? 1:0; // 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] != -1) //记忆化搜索
        return dp[site][mod];
    int len = f == 1?num[site]:9; //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 = 0;
    for(int i = 0; i <= len; i++)
        ans += dfs(site-1,(mod+i)%10,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,0,sizeof(num));
    int l = 0;
    while(digit)
    {
        num[++l] = digit%10;
        digit /= 10;
    }
    return dfs(l,0,1);
}
int main(void)
{
    int t,cnt = 1;
    ll a,b;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld %lld",&a,&b);
        printf("Case #%d: %lld
",cnt++,solve(b)-solve(a-1));
    }
    return 0;
}