URAL - 1057 Amount of Degrees

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 = 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

input	output
15 20 2 2	3

题意：求k个不相同的指数的b进制的和在[x, y]中

思路：数位DP，但要注意的是只有系数为1的情况，所以每位只能是0，1，至于不相同的只要是取1 的时候不不相同，至于0是不变的

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
const int maxn = 50;

int n, m, b, k;
int num[maxn], dp[maxn][maxn];

int dfs(int cur, int cnt, int flag) {
	if (cur == -1)
		return cnt == k;
	if (!flag && dp[cur][cnt] != -1)
		return dp[cur][cnt];
	int end = flag?num[cur]:b-1;
	int ans = 0;
	for (int i = 0; i <= end && i <= 1; i++) 
		ans += dfs(cur-1, cnt+i, flag&&(i==end)); //+1数才会变化
	if (!flag)
		dp[cur][cnt] = ans;
	return ans;
}

int cal(int x) {
	int len = 0;
	while (x) {
		num[len++] = x%b;
		x /= b;
	}
	return dfs(len-1, 0, 1);
}

int main() {
	memset(dp, -1, sizeof(dp));
	while (scanf("%d%d%d%d", &n, &m, &k, &b) != EOF) {
		int ans = cal(m) - cal(n-1);	
		printf("%d\n", ans);
	}
	return 0;
}