Ural Amount of Degrees（数位dp）

传送门

Amount of Degrees

Time limit: 1.0 second
Memory limit: 64 MB

Description

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 = 24+20,
18 = 24+21,
20 = 24+22.

Input

The first line of input contains integers X and Y, separated with a space (1 ≤ X ≤ Y ≤ 231−1). 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

解题思路

题意：

思路：

#include<iostream>
#include<string>
using namespace std;
int f[32][32];

int change(int x, int b)
{
    string s;
    do
    {
        s = char('0' + x % b) + s;
        x /= b;
    }
    while (x > 0);
    for (int i = 0; i < s.size(); ++i)
        if (s[i] > '1')
        {
            for (int j = i; j < s.size(); ++j) s[j] = '1';
            break;
        }
    x = 0;
    for (int i = 0; i < s.size(); ++i)
        x = x | ((s[s.size() - i - 1] - '0') << i);   //或运算，在此相当于加法
    return x;
}

void init()//预处理f
{
    f[0][0] = 1;
    for (int i = 1; i <= 31; ++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 calc(int x, int k) //统计[0..x]内二进制表示含k个1的数的个数
{
    int tot = 0, ans = 0; //tot记录当前路径上已有的1的数量，ans表示答案
    for (int i = 31; i > 0; --i)
    {
        if (x & (1 << i))     //该位上是否为1
        {
            ++tot;
            if (tot > k) break;
            x = x ^ (1 << i);  //将这一位置0
        }
        if ((1 << (i - 1)) <= x)
        {
            ans += f[i - 1][k - tot];
        }
    }
    if (tot + x == k) ++ans;
    return ans;
}

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