hdu4734 F(x) (数位dp)

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
   
   
   
   

 

题意：设n是十进制数x的位数，F(x) = An * 2^(n-1) + An-1 * 2^(n-2) + ... + A2 * 2 + A1 * 1.让你算出0~B中F(y)小于等于F(A)的数的个数。

思路:一开始用dp[i][num][jieguo]表示前i位总和为num，要求的数为jieguo的方案数，但是这样就爆内存了。所以用dp[i][num]表示算到前i位，剩余num，且后面算下去结果大于等于0的方案数，这样就可以转移了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
int dp[11][6000];
int a,b,jieguo;
int wei[12];
int po[20]={1,2,4,8,16,32,64,128,256,512,1024,2048};


int get(int state)
{
    int i,j;
    int shu[12],len=0,sum=0;
    while(state){
        shu[len]=state%10;
        sum+=shu[len]*po[len];
        len++;
        state/=10;
    }
    return sum;
}


int dfs(int pos,int num,int lim,int zero)
{
    int i,j;
    if(num<0)return 0;
    if(pos==-1){
        if(num>=0)return 1;
        return 0;
    }
    if(lim==0 && dp[pos][num]!=-1 ){
        return dp[pos][num];
    }
    int ed=lim?wei[pos]:9;
    int ans=0,num1;
    for(i=0;i<=ed;i++){
        if(zero&&i==0){
            num1=num;
        }
        else{
            num1=num-po[pos]*i;
        }
        ans+=dfs(pos-1,num1,lim&&i==ed,zero&&i==0 );
    }
    if(lim==0){
        dp[pos][num]=ans;
    }
    return ans;
}


int solve(int x)
{
    int i,j,tot=0;
    while(x){
        wei[tot++]=x%10;
        x/=10;
    }
    return dfs(tot-1,jieguo,1,1);
}

int main()
{
    int n,m,i,j,T,cas=0;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&a,&b);
        jieguo=get(a);
        cas++;
        printf("Case #%d: %d\n",cas,solve(b));
    }
    return 0;
}