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


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zhuyuanchen520

 

//31MS    276K    843 B    C++
/*
数位DP：
    好像没什么好讲的... 
*/ 
#include<stdio.h>
#include<string.h>
__int64 dp[20][10];
int digit[20];
void init()    //预处理 
{
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=1;i<20;i++)
        for(int j=0;j<10;j++)
            for(int k=0;k<10;k++)
                dp[i][(j+k)%10]+=dp[i-1][k]; 
}
__int64 deal(__int64 n) //求0~n-1的符合条件的数 
{
    int m=0,last=0;
    __int64 ans=0;
    while(n){
        digit[++m]=n%10;
        n/=10;
    }
    for(int i=m;i>=1;i--){
        for(int j=0;j<digit[i];j++)
            ans+=dp[i-1][(j+last)%10];
        last=(last+digit[i])%10;
    }
    //printf("*%I64d\n",ans);
    return ans-1;
}
int main(void)
{
    int t,k=1;
    __int64 a,b;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&a,&b);
        printf("Case #%d: %I64d\n",k++,deal(b+1)-deal(a));
    }
    return 0;
}