ural1057 Amount of degrees

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

简单的DFS，记忆化搜索。

直接上代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int digit[20];
int k;
long long dp[40][20][25];

long long DFS(int pos,int num,int b,bool inf)
{
    int i,j;
    if (pos==-1) return (num==k);
    if (num>k) return 0;
    if (!inf && dp[pos][num][b]!=-1) return dp[pos][num][b];
    int end=inf ? min(1,digit[pos]) : 1;
    long long ans=0;
    for (i=0;i<=end;i++)
    {
        if (i==0) ans+=DFS(pos-1,num,b,inf && (i==digit[pos]));
        else ans+=DFS(pos-1,num+1,b,inf && (i==digit[pos]));
    }
    if (!inf)
    {
        dp[pos][num][b]=ans;
    }
    return ans;
}

long long Calc(int t,int b)
{
    int pos=0;
    while(t)
    {
        digit[pos++]=t%b;
        t/=b;
    }
    return DFS(pos-1,0,b,1);
}

int main()
{
    int i,j,n,b;
    int x,y;
    memset(dp,-1,sizeof(dp));
    scanf("%d%d%d%d",&x,&y,&k,&b);
    printf("%lld\n",Calc(y,b)-Calc(x-1,b));
    return 0;
}