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

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 ;

 }

巴特西

HDU 4722 数位dp

Good Numbers

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