HDOJ 4389 X mod f(x)

数位DP........

X mod f(x)

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1403    Accepted Submission(s): 599

Problem Description

Here is a function f(x):
　　　int f ( int x ) {
　　　 if ( x == 0 ) return 0;
　　　 return f ( x / 10 ) + x % 10;
　　　}

　　　Now, you want to know, in a given interval [A, B] (1 <= A <= B <= 10 ⁹), how many integer x that mod f(x) equal to 0.

 

Input

　　　The first line has an integer T (1 <= T <= 50), indicate the number of test cases.
　　　Each test case has two integers A, B.

 

Output

　　　For each test case, output only one line containing the case number and an integer indicated the number of x.

 

Sample Input

2
1 10
11 20

 

Sample Output

Case 1: 10
Case 2: 3

 

Author

WHU

 

Source

2012 Multi-University Training Contest 9

 

Recommend

zhuyuanchen520

 

 

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; int bit[ 10],dp[ 10][ 82][ 82][ 82]; int dfs( int pos, int sum, int mod, int res, bool limit) {      if(mod> 81||sum>mod)  return  0;      if(pos==- 1)  return sum==mod&&res== 0;      if(!limit&&~dp[pos][sum][mod][res])          return dp[pos][sum][mod][res];      int end=limit?bit[pos]: 9,ans= 0;      for( int i= 0;i<=end;i++)     {         ans+=dfs(pos- 1,sum+i,mod,(res* 10+i)%mod,limit&&i==end);     }      if(!limit)         dp[pos][sum][mod][res]=ans;      return ans; } int calu( int x) {      int pos= 0,ans= 0;;      while(x)     {         bit[pos++]=x% 10;         x/= 10;     }      for( int i= 1;i<=pos* 9;i++)     {         ans+=dfs(pos- 1, 0,i, 0, true);     }      return ans; } int main() {      int t,a,b,cas= 1;     memset(dp,- 1, sizeof(dp));     scanf( "%d",&t);      while(t--)     {         scanf( "%d%d",&a,&b);         printf( "Case %d: %d\n",cas++,calu(b)-calu(a- 1));     }      return  0; }

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