HDU 4734 F(x)

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 768    Accepted Submission(s): 296

 

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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#include <iostream>

#include <cstring>

#include <cstdio>

#include <cmath>

#define N 5000

#define M 10

using namespace std;

int dp[M][M][N];

int a[M],b[M];

int main()

{

    a[1]=1;

    for(int i=2;i<=9;i++)

    {

        a[i] = a[i-1]*2;

    }

    memset(dp,0,sizeof(dp));

    for(int i=0;i<=9;i++)

    {

        for(int j=i;j<=4608;j++)

        {

            dp[1][i][j]=1;

        }

    }

    for(int i=2;i<=9;i++)

    {

        for(int j=0;j<=9;j++)

        {

            for(int x=a[i]*j;x<=4608;x++)

            {

                for(int y=0;y<=9;y++)

                {

                    dp[i][j][x]+=(dp[i-1][y][x-a[i]*j]);

                }

            }

        }

    }

    int t,cas=1;

    scanf("%d",&t);

    while(t--)

    {

        int n,m;

        scanf("%d %d",&n,&m);

        int s=0,base=1;

        while(n!=0)

        {

            s+=((n%10)*base);

            base=base*2;

            n=n/10;

        }

        int Top=0,com=0;

        base=1;

        while(m!=0)

        {

            b[Top++] = m%10;

            com+=(b[Top-1]*base);

            base=base*2;

            m = m/10;

        }

        int res =0;

        if(com<=s)

        {

            res++;

        }

        for(int i=Top-1;i>=0;i--)

        {

            int x = b[i];

            for(int j=0;j<=x-1;j++)

            {

                res+=(dp[i+1][j][s]);

            }

            s-=(x*a[i+1]);

            if(s<0)

            {

                break;

            }

        }

        printf("Case #%d: %d\n",cas++,res);

    }

    return 0;

}