HDU 4722 Good Numbers（位数DP）（2013 ACM/ICPC Asia Regional Online ―― Warmup2）

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 <= 10 ¹⁸).

 

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

题目大意：求a~b中，位上所有数字加起来能整除10的数有多少个。

思路：位数DP。dp[i][j][k]表示第i位小于j，除以10余k的数的个数，j=10只是为了i+1的j=1准备的……第一次写这个写得有点挫啊……

PS：有数学方法但是我不会，我只知道接近于对于0~n接近于n/10……

代码（109MS）：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 typedef long long LL;

 const int MAXN = ;

 LL dp[MAXN][MAXN][MAXN];
 //i位，字头小于j，和为k的数一共有多少个
 LL a, b;
 int T;

 void init() {
     dp[][][] = ;
     for(int i = ; i <= ; ++i) {
         for(int k = ; k <= ; ++k) dp[i][][k] = dp[i - ][][k];
         for(int j = ; j <= ; ++j) {
             for(int k = ; k <= ; ++k)
                 dp[i][j][k] = dp[i][][(k +  - (j - )) % ] + dp[i][j - ][k];
         }
     }
 }

 int t[MAXN];

 LL calculate(LL n) {
     int cnt = ;
     while(n) t[++cnt] = n % , n /= ;
     int sum = ;
     LL ret = ;
     for(int i = cnt; i > ; --i) {
         ret += dp[i][t[i]][( - sum) % ];
         sum = (sum + t[i]) % ;
     }
     return ret;
 }

 int main() {
     ios::sync_with_stdio(false);
     cin>>T;
     init();
     for(int t = ; t <= T; ++t) {
         cin>>a>>b;
         cout<<"Case #"<<t<<": "<<calculate(b + ) - calculate(a)<<endl;
     }
 }