Timus OJ 1057 数位dp

http://acm.timus.ru/problem.aspx?space=1&num=1057

1057. Amount of Degrees

Time limit: 1.0 second
Memory limit: 64 MB

Create a code to determine the amount of integers, lying in the set [X;Y] and being a sum of exactlyK different integer degrees of B.

Example. Let X=15, Y=20, K=2, B=2. By this example 3 numbers are the sum of exactly two integer degrees of number 2:

17 = 24+20,
18 = 24+21,
20 = 24+22.

Input

The first line of input contains integers X and Y, separated with a space (1 ≤ X ≤ Y ≤ 231−1). The next two lines contain integers K and B (1 ≤ K ≤ 20; 2 ≤ B ≤ 10).

Output

Output should contain a single integer — the amount of integers, lying between X and Y, being a sum of exactly K different integer degrees of B.

Sample

inputoutput

15 2022

3

/**Timus OJ 1057 数位dp题目大意：求出在给定区间内由多少个数可以表示为k个不同的b的幂之和解题思路：对于一个数n，可以求比它小的数的个数有多少个满足条件，首先将n转化为b进制，然后用二进制表示状态，如果b进制下第i位上的数为1，那么对应二进制数为1，           如果为0对应二进制位为0，如果b进制下第i位上的数大于1，那么从第i为往后的二进制位全部置1，得到一个二进制数ans那么该问题就转化为求所有小于等于ans           的二进制数中含有m个1的数有多少个？dp[i][j]表示i二进制位数含j个1的数有多少个，采用记忆化搜索写挺方便*/#include <stdio.h>#include <algorithm>#include <string.h>#include <iostream>using namespace std;int x,y,k,b;int bit[35],dp[35][65];int dfs(int len,int num,int flag,int first){    if(len<0)return num==k;    if(flag==0&&dp[len][num]!=-1)        return dp[len][num];    int ans=0;    int end=flag?bit[len]:1;    for(int i=0;i<=end;i++)    {        int t=first&&(i==0);        ans+=dfs(len-1,t?0:num+(i==1),flag&&i==end,t);    }    if(flag==0)        dp[len][num]=ans;    return ans;}int solve(int n){    int len=0;    while(n)    {        bit[len++]=n%b;        n/=b;    }    int ans=0;    for(int i=len-1;i>=0;i--)    {        if(bit[i]>1)        {            for(int j=i;j>=0;j--)                ans|=(1<<j);        }        else        {            ans|=(bit[i]<<i);        }    }    len=0;    while(ans)    {        bit[len++]=ans&1;        ans>>=1;    }    return dfs(len-1,0,1,1);}int main(){    while(~scanf("%d%d%d%d",&x,&y,&k,&b))    {        memset(dp,-1,sizeof(dp));        printf("%d\n",solve(y)-solve(x-1));    }    return 0;}