HDU - 4734 F(x) （数位dp）

Posted 2021-02-15 zgqblogs

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).

InputThe 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)OutputFor 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

 

题意：给定了一个数字权值计算公式，问0-b中有多少数字f(x)<f(a);

思路：

dp[pos][num]表示在数位pos之后，f(x)小于等于num的数量

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int dp[12][5000];
int dig[12];
int fa;
int dfs(int pos,int num,bool limit){
    if(num<0){ return  0;}
    if(pos==-1){
        return  num>=0;
    }
    if(!limit&&dp[pos][num]!=-1){ return dp[pos][num];}
    int up=limit?dig[pos]:9;
    int ans=0;
    for(int i=0;i<=up;i++){
        ans+=dfs(pos-1,num-i*(1<<pos),limit&&i==up);
    }
    if(!limit){dp[pos][num]=ans;}
    return  ans;
}

int solve(int x){
    int pos=0;
    while (x){
        dig[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,fa,true);
}

int cal(int x){
    int tmp=1,ans=0;
    while (x){
        ans+=(x%10)*tmp;
        x/=10;
        tmp<<=1;
    }
    return  ans;
}
int main()
{
//    ios::sync_with_stdio(false);
//    freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    int cases=0;
    memset(dp,-1,sizeof(dp));
    while (T--){

        cases++;
        int a,b;
        scanf("%d%d",&a,&b);
        fa=cal(a);
        printf("Case #%d: %d
",cases,solve(b));
    }
    return 0;
}

View Code