hdu 4722 Good Numbers

Good Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1115 Accepted Submission(s): 414


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


Recommend
zhuyuanchen520

 

 1 //31MS    276K    843 B    C++
 2 /*
 3 数位DP:
 4     好像没什么好讲的... 
 5 */ 
 6 #include<stdio.h>
 7 #include<string.h>
 8 __int64 dp[20][10];
 9 int digit[20];
10 void init()    //预处理 
11 {
12     memset(dp,0,sizeof(dp));
13     dp[0][0]=1;
14     for(int i=1;i<20;i++)
15         for(int j=0;j<10;j++)
16             for(int k=0;k<10;k++)
17                 dp[i][(j+k)%10]+=dp[i-1][k]; 
18 }
19 __int64 deal(__int64 n) //求0~n-1的符合条件的数 
20 {
21     int m=0,last=0;
22     __int64 ans=0;
23     while(n){
24         digit[++m]=n%10;
25         n/=10;
26     }
27     for(int i=m;i>=1;i--){
28         for(int j=0;j<digit[i];j++)
29             ans+=dp[i-1][(j+last)%10];
30         last=(last+digit[i])%10;
31     }
32     //printf("*%I64d\n",ans);
33     return ans-1;
34 }
35 int main(void)
36 {
37     int t,k=1;
38     __int64 a,b;
39     init();
40     scanf("%d",&t);
41     while(t--)
42     {
43         scanf("%I64d%I64d",&a,&b);
44         printf("Case #%d: %I64d\n",k++,deal(b+1)-deal(a));
45     }
46     return 0;
47 }

 

posted @ 2013-10-05 10:06  heaventouch  阅读(165)  评论(0编辑  收藏  举报