hdu4734 数位dp

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3530    Accepted Submission(s): 1317

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)

题意：找出i在0到b之间的f(i)小于等于f(a)的数的个数。

思路：数位dp。主要在于状态转移不好想。dp[i][j]表示i位数比j小的数的个数。用递归完成的话就只需要思考边界和状态转移。

边界：

dp[i][j]如果j小于0，显然是dp[i][j]=0的，如果i==0，说明就是0，显然任何数都比0大，所以dp[i][j]对于j>=0的时候dp[i][j]=1,否则dp[i][j]=0。

状态转移：

dp[i][j]+=dp[i-1][j-k*(1<<(i-1))]；

记忆化:

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
int dp[20][9300];
int digit[20];
int dfs(int pos,int st,bool limit)
{
    if(pos==0)return st>=0;
    if(st<0)return 0;
    if(!limit&&dp[pos][st]!=-1)return dp[pos][st];
    int ans=0;
    int end=limit?digit[pos]:9;
    for(int i=0; i<=end; i++)
        ans+=dfs(pos-1,st-i*(1<<(pos-1)),limit&&(i==end));
    if(!limit)
        dp[pos][st]=ans;
    return ans;
}
int f(int x)
{
    int ans=0;
    int i=0;
    while(x)
        ans+=(x%10)*(1<<(i++)),x/=10;
    return ans;
}
int get(int a,int b)
{
    int bj=0;
    while(b)
        digit[++bj]=b%10,b/=10;
    return dfs(bj,f(a),1);
}
int main()
{
    int t,o=1;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("Case #%d: %d
",o++,get(a,b));
    }
    return 0;
}

非记忆化:dp[i][j][k] 第i位为j小于k的数目

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define K ((1<<9)*9)
using namespace std;

int T, A, B, dp[9][10][K];

inline void init()
{
    memset(dp, 0, sizeof(dp));
    for(int j = 0; j < 10; ++j)
        for(int k = j; k < K; ++k)
            dp[0][j][k] = 1;
    for(int i = 1; i < 9; ++i)
        for(int j = 0; j < 10; ++j)
            for(int k = 0; k < K; ++k)
                for(int l = 0; l < 10 && (k-j*(1<<i)) >= 0; ++l)
                    dp[i][j][k] += dp[i-1][l][k-j*(1<<i)];
}

inline int f(int x)
{
    int ret = 0, i = 1;
    while(x)
    {
        ret += x%10 * i;
        x /= 10;
        i <<= 1;
    }
    return ret;
}

inline int solve(int b, int x)
{
    int w = 0, bb = b, a[20];
    while(bb)
    {
        a[w++] = bb%10;
        bb /= 10;
    }
    int ret = 0;
    if(f(b) <= x) ret++;
    for(int i = w-1; i >= 0; --i)
    {
        for(int j = 0; j < a[i]; ++j)
            ret += dp[i][j][x];
        x -= a[i]*(1<<i);
if(x < 0) break;

    }
    
    return ret;
}

int main(int argc, char **argv)
{
    init();
    scanf("%d", &T);
    for(int cas = 1; cas <= T; ++cas)
    {
        scanf("%d%d", &A, &B);
        printf("Case #%d: %d
", cas, solve(B, f(A)));
    }

    return 0;
}