ural 1057 Amount of Degrees(数位DP)

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 exactly K 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 = 2 4+2 0,
18 = 2 4+2 1,
20 = 2 4+2 2.

Input

The first line of input contains integers  X and  Y, separated with a space (1 ≤  X ≤  Y ≤ 2 31−1). 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

input	output
15 20 2 2	3

Problem Source: Rybinsk State Avia Academy

Tags: none   ( hide tags for unsolved problems )

题意：

首先定义了一种数为特殊数。特殊数只邮k个不同的b^x组成。现在问你[x,y]里有多少这样的特殊数。

思路：

数位DP。dp[i][j]表示二进制表示中。前i为1的个数为j一共有多少个数。其它进制类似处理。注意边界。

详细见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;
int dp[35][35],base[35],bin[35];
void init()
{
    dp[0][0]=base[0]=1;
    for(int i=1;i<=32;i++)
    {
        dp[i][0]=1;
        base[i]=base[i-1]<<1;
        for(int j=1;j<=i;j++)
            dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
    }
}
int calc(int x,int k)
{
    int i,one=0,ans=0;
    for(i=31;i>=0;i--)
    {
        if(x&base[i])
        {
            if(one>k)
                break;
            ans+=dp[i][k-one];
            one++;
            x-=base[i];
        }
    }
    if(one==k)
        ans++;
    return ans;
}
int getbin(int x,int b)
{
    int ct=0,i,ret=0;
    if(!x)
        return x;
    while(x)
    {
        bin[ct++]=x%b;
        x/=b;
    }
    for(i=ct-1;i>=0;i--)
    {
        if(bin[i]>1)
            break;
        if(bin[i])
            ret=ret<<1|1;
        else
            ret<<=1;
    }
    while(i>=0)
    {
        ret=ret<<1|1;
        i--;
    }
    return ret;
}
int main()
{
    int x,y,k,b;

    init();
    while(~scanf("%d%d%d%d",&x,&y,&k,&b))
    {
        y=getbin(y,b);
        x=getbin(x-1,b);
        printf("%d\n",calc(y,k)-calc(x,k));
    }
    return 0;
}