hdu4734 F(x)(数位dp）

题目传送门

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8901    Accepted Submission(s): 3503

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

3 0 100 1 10 5 100

 

Sample Output

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

 

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 

Recommend

 

题意：定义F(x),求在[0,m]中F[x]小于F(n)的数的个数

题解：数位dp

代码：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
int T;
int n,m;
int bit[10];
int dp[10][4600];
int F(int x)
{
    int cnt=0;
    int len=0;
    while(x)
    {
        cnt+=(x%10)*(1<<len);
        x/=10;
        len++;
    }
    return cnt;
}
int dfs(int pos,int sta,bool limit)
{
    if(pos==-1) return sta>=0;
    if(sta<0) return 0;
    if(!limit&&dp[pos][sta]!=-1) return dp[pos][sta];
    int ans=0;
    int up=limit?bit[pos]:9;
    for(int i=0;i<=up;i++)
        ans+=dfs(pos-1,sta-i*(1<<pos),limit&&i==up);
    if(!limit) dp[pos][sta]=ans;
    return ans;
}
int calc(int x)
{
    int len=0;
    while(x)
    {
        bit[len++]=x%10;
        x/=10;
    }
    return dfs(len-1,F(n),true);
}
int main()
{
    scanf("%d",&T);
    int ncase=0;
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        scanf("%d %d",&n,&m);
        printf("Case #%d: ",++ncase);
        printf("%d\n",calc(m));
    }
    return 0;
}