hdu 4734 F(x)(数位dp)

 

Problem Description

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

 

 

Sample Input

3

0 100

1 10

5 100 

 

Sample Output

Case #1: 1

Case #2: 2 Case #3: 13

 

题意:题目给了个f(x)的定义：F(x) = An * 2n-1 + An-1 * 2n-2 + … + A2 * 2 + A1 * 1，Ai是十进制数位，然后给出a,b求区间[0,b]内满足f(i)==f(a)的i的个数。

思路:最开始的思路是dp[len][sum] 然后每次都memset果断tle 然后觉的开三维 空间又过不了 看聚聚的题解知道可以用总和的思想来减去当前的权值

仔细想想这个状态是与f(a)无关的，一个状态只有在sum=0时才满足，如果我们按常规的思想求f(i)的话，那么最后sum=f(a)才是满足的条件。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[2][2]={1,0 ,0,1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
int bits[10];
int dp[10][10007]; //len位 需要凑sum的权值和 
int fa;
int v[11]={1,2,4,8,16,32,64,128,256,512};
int dfs(int len,int sum,bool ismax){
    if(!len) return sum<=fa;
    if(sum>fa) return 0;
    if(!ismax&&dp[len][fa-sum]>=0) return dp[len][fa-sum];
    int up=ismax?bits[len]:9;
    int cnt=0;
    for(int i=0;i<=up;i++){
        cnt+=dfs(len-1,sum+i*v[len-1],ismax&&i==up);
    }
    if(!ismax) dp[len][fa-sum]=cnt;
    return cnt;
}
int solve(int x){
    int len=0;
    while(x){
        bits[++len]=x%10;
        x/=10;
    }
    return dfs(len,0,true);
}
void cal(int a){
    int now=1;
    while(a){
        fa+=now*(a%10);
        a/=10; now*=2;
    }
}
int main(){
    int t;
    scanf("%d",&t);
    int w=0;
    memset(dp,-1,sizeof(dp));
    while(t--){
        fa=0;    
        int a,b; 
        scanf("%d%d",&a,&b);
        cal(a);
        printf("Case #%d: %d\n",++w,solve(b));
    }
    return 0;
}