HDU-4734 F(x)

最新推荐文章于 2021-11-17 16:57:55 发布

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本文详细解析了HDU 4734问题，该问题要求计算在给定区间内，有多少个数字的特定权重不超过预设值。通过使用深度优先搜索和动态规划的方法，文章提供了完整的代码实现，并解释了状态转移方程，帮助读者理解如何高效地解决这类计数问题。

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

Problem Description

For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-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.

题意：按照题中F（x）定义，求[0,B]有多少数小于F（A）

思路：状态设为（第pos位，还未填的位最多还能加权填多少），不能是已经填了多少，否则A一变，dp数组就不再适用了。

#include<bits/stdc++.h>
using namespace std;

int T,A,B,all;
int a[20],dp[20][1<<15];

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

int solve(int n)
{
	int pos=0;
	while(n)
	{
		a[pos++]=n%10;
		n/=10;
	}
	return dfs(pos-1,0,1);
}
int F(int n)
{
	if(!n)return 0;
	int ret=F(n/10);
	return ret*2+n%10;
}

int main()
{
//	freopen("input.in","r",stdin);
	cin>>T;
	memset(dp,-1,sizeof(dp));
	for(int i=1;i<=T;i++)
	{
		cin>>A>>B;
		all=F(A);
		cout<<"Case #"<<i<<": "<<solve(B)<<endl;
	}
		
	return 0;
}