HDU 4722 (数位DP 水~)

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3814    Accepted Submission(s): 1213

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.
 

 

求出区间里面每个数位和模10余0的个数.

#include <bits/stdc++.h>using namespace std;#define mod 1000000007#define maxn 22typedef long long ull;ull a, b;int bit[maxn], l;ull dp[maxn][11];ull dfs (int pos, int m1, bool f2) {    //当前的位数 之前模10余数 是不是可以取到9    if (pos == 0) {        return m1 == 0;    }    if (f2 && dp[pos][m1] != -1) {        return dp[pos][m1];    }    ull Max = (f2 ? 9 : bit[pos]);    ull ans = 0;    for (int i = 0; i <= Max; i++) {        ans += dfs (pos-1, (m1+i)%10, f2 || (i<Max));    }    if (f2)        dp[pos][m1] = ans;    return ans;}ull f (ull num) {    l = 0;    while (num) {        bit[++l] = num%10;        num /= 10;    }    return dfs (l, 0, 0);}int main () {    //freopen ("in.txt", "r", stdin);    int t, kase = 0;    memset (dp, -1, sizeof dp);    cin >> t;    while (t--) {        cin >> a >> b;        printf ("Case #%d: %lld\n", ++kase, f (b)-f (a-1));    }    return 0;}