ural 1057. Amount of Degrees(数位DP，4级)

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本文介绍了一个算法问题，即确定在指定区间[X;Y]内能由K个不同B的幂次和组成的整数数量。通过转换为B进制表示，并使用动态规划方法解决该问题。

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

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Difficulty: 767 Printable version Submit solution Discussion (36)
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思路：转成B进制，只要是B进制打头的是01皆可以。DP之

#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=65;
int f[mm][mm];
int bit[mm];
int DP(int n,int num,int b)
{
    int pos=0;
    while(n)
    {
        bit[++pos]=n%b;n/=b;
    }
    int ret=0,one=0;
    for(int i=pos;i>=1;--i)
    {
        if(bit[i]>1)
        {ret+=f[i][num-one];break;//已经大于了
        }
        else if(bit[i])
        {
            ret+=f[i-1][num-one];++one;//相等则下一层可选，多一个1
        }
        if(i==1&&one==num)++ret;//相等的情况
    }
    return ret;
}
int main()
{
    FOR(i,0,mm-1)f[i][0]=1;
    FOR(i,1,mm-1)FOR(j,1,mm-1)
    f[i][j]=f[i-1][j-1]+f[i-1][j];//0 or 1
    int X,Y,K,B;
    while(~scanf("%d%d%d%d",&X,&Y,&K,&B))
    {
        printf("%d\n",DP(Y,K,B)-DP(X-1,K,B));
    }
    return 0;
}