ural 1057 Amount of degrees-数位统计-数位dp

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 betweenX and Y, being a sum of exactly K different integer degrees of B.

Example

input	output
15 20 2 2	3

题目大意：给定区间【x,y】中满足区间中的数正好满足是由k个互不相等的B的整数次幂之和，求出有几个这样子的数

思路：http://wenku.baidu.com/view/d2414ffe04a1b0717fd5dda8.html

观察一下上面的网页吧，很仔细了

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int x,y,k,b;
int c[33][33];
void Init() //其实就是组合数
{
    c[0][0]=1;
    for(int i=1; i<=31; i++)
    {
        c[i][0]=c[i-1][0];
        for(int j=1; j<=i; j++)
            c[i][j]=c[i-1][j]+c[i-1][j-1];
    }
}

int change(int n) //转换进制问题了
{
    int w[33];
    int len=0;
    while(n)
    {
        w[len++]=n%b;
        n/=b;
    }
    int ans=0;
    for(int i=len-1; i>=0; i--)
        if(w[i]>1)
        {
            for(int j=i;j>=0;j--)
                ans+=(1<<j);
            break;
        }
        else
            ans+=(w[i]<<i);
        return ans;
}

int calc(int x,int k)
{
    int tot=0,ans=0;
    for(int i=31;i>=0;i--)
    {
        if(x&(1<<i))
        {
            tot++;
            if(tot>k) break;
            x=x^(1<<i);
        }
        if(1<<(i-1)<=x)
            ans+=c[i-1][k-tot]; //第i位为1（从0开始的）,那么后面还剩下i个数字,后面的第一个数字为0，
                                 //从i-1个数字中任意挑k-tot个
    }

    if(tot+x==k)
        ans++;
    return ans;
}

int main()
{
    Init();
    cin>>x>>y>>k>>b;
    cout<<calc(change(y),k)-calc(change(x-1),k)<<endl;
    return 0;
}