HDU4734 F(x)

最新推荐文章于 2021-01-19 21:31:29 发布

Jelly_acm

最新推荐文章于 2021-01-19 21:31:29 发布

阅读量374

点赞数

分类专栏： HDU ----数位dp

本文链接：https://blog.csdn.net/Jelly_acm/article/details/77926460

版权

HDU 同时被 2 个专栏收录

181 篇文章

订阅专栏

----数位dp

7 篇文章

订阅专栏

F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6383 Accepted Submission(s): 2456

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

————————————————————————————————

题目的意思求0到B 有多少个经过计算F(x)不大于F(A)的有几个

思路：数位dp，2位数组表示到第len位为止，小于等于sum有多少个，dfs求解

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 10000005
#define Mod 10001
using namespace std;
#define LL long long

int dp[20][30000];
int a[100];

int dfs(int len,int sum,bool limit)
{
    if(len<0)
        return sum>=0;
    if(sum<0)
        return 0;
    if(dp[len][sum]!=-1&&!limit)
        return dp[len][sum];
    int up=limit?a[len]:9;
    int ans=0;
    for(int i=0; i<=up; i++)
    {
        ans+=dfs(len-1,sum-i*(1<<len),limit&&i==up);
    }
    return limit?ans:dp[len][sum]=ans;
}

int solve(int y,int x)
{
    int sum=0;
    int tmp=1;
    while(y>0)
    {
        sum+=y%10*tmp;
        y/=10;
        tmp*=2;
    }

    int cnt=0;
    while(x>0)
    {
        a[cnt++]=x%10;
        x/=10;
    }
    return dfs(cnt-1,sum,1);
}

int main()
{
    int n,m;
    int T;
    int q=1;
    memset(dp,-1,sizeof dp);
    for(scanf("%d",&T); T--;)
    {
        scanf("%d%d",&m,&n);
        printf("Case #%d: %d\n",q++,solve(m,n));
    }
    return 0;
}