~~~~2013成都网赛f(x) hdu4734 数位dp~~~~_with n digits (anan-1an-2 ... a2a1), we define its-优快云博客

本文链接：https://blog.youkuaiyun.com/w750636248/article/details/11933789

本文介绍了一个数值权重计算问题，即对于一个十进制数x，定义其权重F(x)，并探讨了如何计算在0到B范围内，权重不超过F(A)的数的数量。文章通过预处理方法，使用动态规划技巧来高效解决此问题。

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F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 712 Accepted Submission(s): 269

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

Source

2013 ACM/ICPC Asia Regional Chengdu Online

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#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#define N 100010
#define eps (1e-8)
#define INF 99999999
using namespace std;
int c[10][9*512];//c[i][j]表示有i位且f(x)<=j的x个数;
void init()
{
    int i,j,k;
    for(i=0;i<10;i++)
        c[i][0]=1;//开始忘记了初始化这个了，WA了好久。。。哎
    for(i=0;i<9*512;i++)
        c[0][i]=1;
    for(i=0;i<9*512;i++)
        c[1][i]=min(10,i+1);
    for(i=2;i<=9;i++)
        for(j=1;j<9*512;j++)
        {
            for(k=0;k<=9 && j-k*(1<<(i-1))>=0;k++)
                    c[i][j]+=c[i-1][j-k*(1<<(i-1))];
        }
}
int f(int x)
{
    int ans=0,q=1;
    while(x)
    {
        ans+=q*(x%10);
        q*=2;
        x/=10;
    }
    return ans;
}
int solve(int a,int b)
{
    int s[11],d=0,ans=0;
    if(a==0) return 1;
    while(b)
    {
        s[d++]=b%10;
        b/=10;
    }
    for(d--;d>=0;d--)
    {
        for(int i=0;i<s[d];i++)
        {
            if(a-i*(1<<d)<0) return ans;
            ans+=c[d][a-i*(1<<d)];
        }
        a-=s[d]*(1<<d);
        if(a<0) return ans;
    }
    return ans;
}
int main()
{
    int t,cas=1;
    init();
    scanf("%d",&t);
    while(t--)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        printf("Case #%d: %d\n",cas++,solve(f(a),b+1));
    }
    return 0;
}

Run ID

Submit Time

Judge Status

Pro.ID

Exe.Time

Exe.Memory

Code Len.

Language

9214984

2013-09-23 15:27:17

Accepted

4734