HDU 4722 数位dp

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4753    Accepted Submission(s): 1511

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：对一个数字 每一位求和%10==0 问一个区间内有多少个这类数字

 题解：模板题

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#define ll  __int64
#define dazhi 2147483647
#define bug() printf("!!!!!!!")
#define M 200005
using namespace  std;
int t;
ll l,r;
ll bit[20];
ll dp[100][15];
ll functi(int pos ,int flag,int m)
{
    if(pos==0) return (m==0);
    if(flag&&dp[pos][m]!=-1) return dp[pos][m];
    ll x=flag?9:bit[pos];
    ll ans=0;
    for(ll i=0;i<=x;i++)
        ans+=functi(pos-1,flag||i<x,(m+i)%10);
    if(flag) dp[pos][m]=ans;
    return ans;
}
ll fun(ll num)
{
    int len=0;
    memset(bit,0,sizeof(bit));
    while(num)
    {
        bit[++len]=num%10;
        num/=10;
    }
    return functi(len,0,0);
}
int main()
{
    while(scanf("%d",&t)!=EOF)
    {
        memset(dp,-1,sizeof(dp));
         for(int i=1;i<=t;i++)
         {
             scanf("%I64d %I64d",&l,&r);
             printf("Case #%d: %I64d
",i,fun(r)-fun(l-1));
         }
    }
    return 0;
}