hdu 4734 F(x)(数位DP，4级)-CSDN博客

本文链接：https://blog.csdn.net/nealgavin/article/details/11703049

F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 443 Accepted Submission(s): 160

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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liuyiding

数位DP：长度为x,值小于y的最优值可以满足所有，因此不用重复初始化

#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define ll(x) (1<<x)
using namespace std;
int dp[13][9*ll(11)];///长度为x,<y的个数值
int bit[13],A,B,low[13];
int F(int x)
{ int ret=0,z=1;
  while(x)
  {
    ret+=x%10*z;z<<=1;x/=10;
  }
  return ret;
}
int DP(int pp,int st,bool big)
{
  if(pp==0)return st>=0;
  if(big&&dp[pp][st]!=-1)return dp[pp][st];
  int kn=big?9:bit[pp];
  int ret=0;
  FOR(i,0,kn)
  {
    ret+=DP(pp-1,st-i*low[pp-1],big||kn!=i);
  }
  if(big)dp[pp][st]=ret;
  return ret;
}
int get(int a,int b)
{ int pos=0;
  while(b)
  {
    bit[++pos]=b%10;b/=10;
  }
  return DP(pos,F(a),0);
}
int main()
{ low[0]=1;
  FOR(i,1,10)low[i]=low[i-1]*2;
  int cas;clr(dp,-1);
  while(~scanf("%d",&cas))
  {
    FOR(ca,1,cas)
    {
      scanf("%d%d",&A,&B);
      printf("Case #%d: ",ca);
      printf("%d\n",get(A,B));
    }
  }
}