HDU 4374--F(x)

F(x)

Time Limit:1  Seconds    Memory Limit:32  MB    

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

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

题目意思很简单，按要求计算0-B之间的F（x）值，算小于等于F（A）的个数。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;

int dp[11][10001];  //dp[i][j] ..表示i位数 <= j的个数
int bit[11];   //记录每一位

int F(int x)
{
    int sum = 0;
    int mul = 0;
    while(x)
    {
        int y = (x % 10) * (1 << mul);
        sum += y;
        mul++;
        x /= 10;
    }
    return sum;
}

int A,B;
int t;

int dfs(int pos,int num,int full)
{
    if(pos == -1)return num >= 0;                         //求到最低位做出判断
    if(num < 0)return 0;
    if(!full && dp[pos][num] != -1)return dp[pos][num];  //返回记忆化的值
    int end = full?bit[pos]:9;                             //取该位的最大值
    int ans = 0;
    for(int i = 0;i <= end;i++)
    {
        ans += dfs(pos - 1,num - i * (1 << (pos)) ,full && i == end);
    }
    if(!full)dp[pos][num] = ans;                         //记忆化记录
    return ans;
}

int Cal()
{
    int y = B;
    int cnt = 0;
    while(y)
    {
        bit[cnt++] = y % 10;
        y /= 10;
    }
    return dfs(cnt - 1,F(A),1);
}

int main()
{
    scanf("%d",&t);
    int cas = 1;
    memset(dp,0xff,sizeof(dp));
    while(t--)
    {    
        scanf("%d%d",&A,&B);
        printf("Case #%d: %d
",cas++,Cal());
    }
    return 0;
}