hdu 4734 数位dp

最新推荐文章于 2022-09-22 09:47:31 发布

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最新推荐文章于 2022-09-22 09:47:31 发布

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分类专栏： DP 数位dp

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本文介绍了一种使用数位动态规划（数位DP）的方法来解决特定的重量计算问题。具体而言，对于给定的两个数A和B，我们需要找出在0到B范围内，有多少个数的权重不超过数A的权重。通过记忆化搜索优化递归过程，有效地解决了这个问题。

http://acm.hdu.edu.cn/showproblem.php?pid=4734

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

Sample Output

  
   Case #1: 1
Case #2: 2
Case #3: 13

/**
hdu 4734  数位dp 记忆化搜索
题目大意：求给定区间1~B中所有小于f（A）的数。
解题思路：记忆化搜索， 标记dp[i][j]表示i位数比j小的数的个数。时间卡的很紧，不用记忆化会超时的
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

int A,B,dp[11][50000],a[25];
int len;

int f(int n)
{
    int ans=0,len=1;
    while(n)
    {
        ans+=n%10*len;
        len*=2;
        n/=10;
    }
   /// printf("f(a)=%d\n",ans);
    return ans;
}

int dfs(int len,int ans,int flag)
{
    if(len<0) return ans>=0;
    if(ans<0) return 0;
    int sum=0;
    if(!flag&&dp[len][ans]!=-1)return dp[len][ans];
    int end=flag?a[len]:9;
    for(int i=0; i<=end; i++)
    {
        sum+=dfs(len-1,ans-i*(1<<len),flag&&i==end);
    }
    if(!flag)dp[len][ans]=sum;
    return sum;
}
int main()
{
    int T,tt=0;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        scanf("%d%d",&A,&B);
        len=0;
        while(B)
        {
            a[len++]=B%10;
           /// printf("%d ",a[len-1]);
            B/=10;
        }
        ///printf("\n");
        printf("Case #%d: %d\n",++tt,dfs(len-1,f(A),1));
    }
    return 0;
}