hud 4722 Good Numbers(数位 DP)

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

21 101 20

 

Sample Output

Case #1: 0Case #2: 1
Hint
The answer maybe very large, we recommend you to use long long instead of int.
 

 

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;#define LL long longLL digit[30];LL dp[30][30]; //位数为i时，和为j时满足条件的个数LL dfs(LL pos, LL pre, bool doing){    if(pos == -1) return pre == 0;    if(!doing && dp[pos][pre]!=-1)        return dp[pos][pre];    LL ans = 0, npre;    LL end = doing ? digit[pos] : 9;//doing 是否为真为真end=digit[pos]    for(LL i = 0 ; i <= end ; i ++)    {        npre = (pre + i) % 10;        ans += dfs(pos - 1, npre, doing && (i == end));    }    if(!doing) dp[pos][pre] = ans;    return ans;}LL calc(LL x){    LL pos = 0;    memset(dp, -1,sizeof(dp));    while(x)    {        digit[pos ++]= x % 10;        x /= 10;    }    return dfs(pos - 1, 0, 1);//pos是位数，}LL a,b,ans;int main(){    int T;    scanf("%d",&T);    for(int ncase = 1; ncase <= T; ncase++)    {        scanf("%I64d %I64d",&a,&b);        ans = calc(b) - calc(a - 1);        printf("Case #%d: %I64d\n",ncase,ans);    }    return 0;}