hdu 4734 F(x) （数位dp）

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1140    Accepted Submission(s): 459

Problem Description

For a decimal number x with n digits (A _nA _n-1A _n-2 ... A ₂A ₁), we define its weight as F(x) = A _n * 2 ^n-1 + A _n-1 * 2 ^n-2 + ... + A ₂ * 2 + A ₁ * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).

 

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
For each test case, there are two numbers A and B (0 <= A,B < 10 ⁹)

 

Output

For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer.

 

Sample Input

   
   
   
   

    
    
    
    3
0 100
1 10
5 100
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 1
Case #2: 2
Case #3: 13
   
   
   
   

 

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 

题意：

看懂f(x)函数之后，求[0,B]内的f(x)的值小于f(a)的有多少个。

思路：

数位dp， 由于f(x)的值不大，可以考虑把f(x)放入状态中。

dp[i][j][k] - i 位 j 开头f(x)值为k的数的个数。

那么有dp[i][j][k]=dp[i-1][p][k-j*2^(i-1)] 。

然后算一个区间内的个数时，一位一位往下算就够了。

考虑到时间的原因，采用了一个sum数组纪录前缀和。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot,flag;
int a,b,fac[10];
int dp[10][10][5500],sum[10][10][5500];

void solve()
{
    int i,j,t=0,u,v;
    for(i=9;i>=1;i--)
    {
        t+=((a/fac[i-1])%10)*(1<<i-1);
    }
    a=t;
    ans=0;
    for(i=9;i>=1;i--)
    {
        u=(b/fac[i-1])%10;
        for(j=0;j<u;j++)
        {
            ans+=sum[i][j][a];
        }
        a-=u*(1<<i-1);
        if(a<0) return ;
    }
    ans++;
}
int main()
{
    int i,j,k,p,q,t,test=0;
    fac[0]=1;
    for(i=1;i<10;i++)
    {
        fac[i]=fac[i-1]*10;
    }
    dp[0][0][0]=1;
    for(i=1;i<=9;i++)
    {
        for(j=0;j<=9;j++)
        {
            for(k=0;k<=5000;k++)
            {
                t=0;
                for(p=0;p<=9;p++)
                {
                    if((k-(1<<i-1)*j)>=0) t+=dp[i-1][p][k-(1<<i-1)*j];
                }
                dp[i][j][k]=t;
                if(k>0) sum[i][j][k]=sum[i][j][k-1]+t;
                else sum[i][j][k]=t;
            }
        }
    }
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&a,&b);
        solve();
        printf("Case #%d: %d\n",++test,ans);
    }
    return 0;
}
/*
3
0 100
1 10
5 100
Case #1: 1
Case #2: 2
Case #3: 13
*/