HDU 4734 F(x) （2013成都网络赛，数位DP）

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 382    Accepted Submission(s): 137

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

 

 

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liuyiding

 

 

 

 

数位DP的水题

dp[i][j]表示i位值<=j 的总数

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013/9/14 星期六 12:45:42
File Name     :2013成都网络赛\1007.cpp
************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

int dp[20][200000];

int bit[20];



int dfs(int pos,int num,bool flag)
{
    if(pos == -1)return num >= 0;
    if(num < 0)return 0;
    if(!flag && dp[pos][num] != -1)
        return dp[pos][num];
    int ans = 0;
    int end = flag?bit[pos]:9;
    for(int i = 0;i <= end;i++)
    {

        ans += dfs(pos-1,num - i*(1<<pos),flag && i==end);
    }
    if(!flag)dp[pos][num] = ans;
    return ans;
}

int F(int x)
{
    int ret = 0;
    int len = 0;
    while(x)
    {
        ret += (x%10)*(1<<len);
        len++;
        x /= 10;
    }
    return ret;
}
int A,B;
int calc()
{
    int len = 0;
    while(B)
    {
        bit[len++] = B%10;
        B/=10;
        //cout<<bit[len-1]<<endl;
    }
    //cout<<F(A)<<endl;
    return dfs(len-1,F(A),1);
}




int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    int iCase = 0;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        iCase++;
        scanf("%d%d",&A,&B);
        printf("Case #%d: %d\n",iCase,calc());
    }
    return 0;
}