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 <= 109), 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
 

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;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )

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