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Good Numbers
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.
题目大意：求[ a , b ] 有多少个数是 各个位数的和mod10等于0 的数
思路：比赛的时候想到要用数位dp，但是很久没用有点忘了，没推对
数位dp dp [ i ] [ j ] 为长度为 i 和mod10为 j 的个数
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
const int N=20;
ll dp[20][10];
int d[20];
ll dfs(int len,int k,int lim)
{
if(len==0) return k==0;
if(!lim&&dp[len][k]!=-1) return dp[len][k];
int maxx=lim?d[len]:9;
ll ans=0;
for(int i=0;i<=maxx;i++)
ans+=dfs(len-1,(i+k)%10,lim&&(i==maxx));
if(!lim) dp[len][k]=ans;
return ans;
}
ll solve(ll x)
{
int k=0;
while(x)
{
d[++k]=x%10;
x/=10;
}
return dfs(k,0,1);
}
int main()
{
int T,t=1;
scanf("%d",&T);
memset(dp,-1,sizeof(dp));
while(T--)
{
ll a,b;
scanf("%lld%lld",&a,&b);
ll ans=solve(b)-solve(a-1);
printf("Case #%d: %lld\n",t++,ans);
}
return 0;
}