hdu 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)


[align=left]Problem Description[/align]

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.

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

[align=left]Output[/align]
　　　For each test case, output only one line containing the case number and an integer indicated the number of x.

[align=left]Sample Input[/align]

2
1 10
11 20

[align=left]Sample Output[/align]

Case 1: 10
Case 2: 3

[align=left]Author[/align]
WHU

[align=left]Source[/align]
2012 Multi-University Training Contest 9
dp进阶之路写法；
或者分块打表；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
int f[11][83][83][83],bit[11];
int dp(int pos,int sum,int m,int p,int flag)
{
if(pos==0)return (sum==p&&m==0);
if(flag&&f[pos][sum][m][p]!=-1)return f[pos][sum][m][p];
int x=flag?9:bit[pos];
int ans=0;
for(int i=0;i<=x;i++)
{
ans+=dp(pos-1,sum+i,(m*10+i)%p,p,flag||i<x);
}
if(flag)f[pos][sum][m][p]=ans;
return ans;
}
int getans(int x,int p)
{
int len=0;
while(x)
{
bit[++len]=x%10;
x/=10;
}
return dp(len,0,0,p,0);
}
int main()
{
int T,cas=1;
memset(f,-1,sizeof(f));
scanf("%d",&T);
while(T--)
{
int l,r;
scanf("%d%d",&l,&r);
int ans=0;
for(int i=1;i<=81;i++)
ans+=getans(r,i)-getans(l-1,i);
printf("Case %d: %d\n",cas++,ans);
}
return 0;
}