Ural Amount of Degrees（数位dp）

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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：

 

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#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; }