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Ural Amount of Degrees（数位dp）

2017-04-28 17:41 309 查看

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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

input

output

15 20
2
2

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;
}

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