数位dp入门(Amount of Degrees)

图片来源:https://wenku.baidu.com/view/d2414ffe04a1b0717fd5dda8.html 

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.

 

 

 

#include<bits/stdc++.h>
using namespace std;
int digit[33];
typedef long long LL;
const int N = 50;
int f[N][N]; //f[i][j]表示前i个中选 j 个 1的个数
void init()
{
    f[0][0] =  1;
    for(int i=1;i<33;i++)
	{
        f[i][0] = f[i-1][0];
        for(int j=1;j<=i;j++)
        {
        	f[i][j] = f[i-1][j] + f[i-1][j-1];
		 } 
    }
}
int solve(int x,int k,int B)
{
	int len=-1;
    //vector<int> v;
    while(x)
    {
    	digit[++len]=x%B; 
        //v.push_back(x%B);
        x/=B;
    }
    int cnt = 0,ans = 0;
    for(int i=len;i>=0;i--)
    {
        if(digit[i]==1) // 如果为 1，则依次求解
        {
            ans+=f[i][k-cnt]; //need 
            cnt++;
            if(cnt==k)
                break;
        }
        else if(digit[i]>1) //假如大于1的话，相当于所有的位可以为 1，所以直接求解跳出
        {
            ans += f[i+1][k-cnt];
            break;
        }
    }
    if(cnt==k)
        ans++;
    return ans;
}
int main()
{
    init();
    int x,y,k,B;
    while(~scanf("%d%d%d%d",&x,&y,&k,&B))
    {
        printf("%d\n",solve(y,k,B)-solve(x-1,k,B));
    }
    return 0;
}