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[];
 ll dp[][];
 ll functi(int pos ,int flag,int m)
 {
     if(pos==) return (m==);
     if(flag&&dp[pos][m]!=-) return dp[pos][m];
     ll x=flag?:bit[pos];
     ll ans=;
     for(ll i=;i<=x;i++)
         ans+=functi(pos-,flag||i<x,(m+i)%);
     if(flag) dp[pos][m]=ans;
     return ans;
 }
 ll fun(ll num)
 {
     int len=;
     memset(bit,,sizeof(bit));
     while(num)
     {
         bit[++len]=num%;
         num/=;
     }
     return functi(len,,);
 }
 int main()
 {
     while(scanf("%d",&t)!=EOF)
     {
         memset(dp,-,sizeof(dp));
          for(int i=;i<=t;i++)
          {
              scanf("%I64d %I64d",&l,&r);
              printf("Case #%d: %I64d\n",i,fun(r)-fun(l-));
          }
     }
     return ;
 }