找规律/数位DP HDOJ 4722 Good Numbers

题目传送门

 /*
     找规律/数位DP：我做的时候差一点做出来了，只是不知道最后的 is_one ()
                     http://www.cnblogs.com/crazyapple/p/3315436.html
     数位DP：http://blog.csdn.net/libin56842/article/details/11580497
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <cmath>
 using namespace std;

 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;

 int is_one(long long n)
 {
     long long m = n;
     long long i = n /  * ;

     for (; i<=m; ++i)
     {
         long long tmp = i;    int sum = ;
         while (tmp)
         {
             sum += tmp % ;
             tmp /= ;
         }
         if (sum %  == )    return ;
     }

     return ;
 }

 long long get_num(long long n)
 {
     if (n < )    return ;
     if (n <= )    return ;

     return n/ + is_one (n);
 }

 int main(void)        //HDOJ 4722 Good Numbers
 {
     //freopen ("G.in", "r", stdin);

     int t, cas = ;
     long long l, r;
     scanf ("%d", &t);
     while (t--)
     {
         scanf ("%I64d%I64d", &l, &r);

         printf ("Case #%d: %I64d\n", ++cas, get_num (r) - get_num (l-));
     }

     return ;
 }

 /*
 Case #1: 0
 Case #2: 1
 Case #3: 1
 */

 /*
     数位DP：基础题，对于一个数，把它每位数分出来，从最高位开始枚举。dp[i][(j+k)%10] += dp[i+1][j];
             dp[i][j]表示前i位数和取模是j的个数
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 typedef long long ll;
 const int MAXN = ;
 const int INF = 0x3f3f3f3f;
 ll dp[MAXN][MAXN];
 int dig[MAXN];

 ll work(ll n)   {
     ll tmp = n; int len = ;
     memset (dig, , sizeof (dig));
     while (tmp) {
         dig[++len] = tmp % ;
         tmp /= ;
     }
     memset (dp, , sizeof (dp));    int x = ;
     for (int i=len; i>=; --i)  {
         for (int j=; j<; ++j)    {
             for (int k=; k<; ++k)    {
                 dp[i][(j+k)%] += dp[i+][j];
             }
         }
         for (int j=; j<dig[i]; ++j)    dp[i][(x+j)%]++;
         x = (x + dig[i]) % ;
     }
     if (!x) dp[][]++;
     return dp[][];
 }

 int main(void)  {       //HDOJ 4722 Good Numbers
     int T, cas = ;  scanf ("%d", &T);
     while (T--) {
         ll l, r;    scanf ("%I64d%I64d", &l, &r);
         printf ("Case #%d: %I64d\n", ++cas, work (r) - work (l-));
     }

     return ;
 }

数位DP