HDOJ 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): 750    Accepted Submission(s): 286

Problem Description

For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-2 + ... + A₂ * 2 + A₁ * 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 < 10⁹)

 

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

30 100
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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#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long int LL;

LL dp[12][111111];
int bit[12];

int dfs(int pos,int sum,bool limit)
{
    if(pos==-1) return 1;
    if(~dp[pos][sum]&&limit==false) return dp[pos][sum];
    int end=limit?bit[pos]:9;
    int res=0;
    for(int i=0;i<=end;i++)
    {
        if((sum-i*(1<<pos))>=0)
            res+=dfs(pos-1,sum-i*(1<<pos),limit&&i==end);
    }
    if(!limit)
        dp[pos][sum]=res;
    return res;
}

int getsum(int x)
{
    int l=1,sum=0;
    while(x)
    {
        sum+=l*(x%10);
        x/=10; l=l*2;
    }
    return sum;
}

LL colu(int x,int y)
{
    int pos=0,sum=getsum(y);
    while(x)
    {
        bit[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,sum,true);
}

int main()
{
    int cas=1,x,y,t;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&y,&x);
        printf("Case #%d: %I64d\n",cas++,colu(x,y));
    }
    return 0;
}

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