[TimusOJ1057]Amount of Degrees

[TimusOJ1057]Amount of Degrees

试题描述

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.

输入

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).

输出

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.

输入示例

15 20
2
2

输出示例

3

数据规模及约定

见“输入”

题解

就是看转换成 b 进制之后是不是只有 0 和 1 并且 1 的个数恰好等于 k。

行了，明白了题意，就没啥可说的了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 35
#define maxk 25
int f[maxn][2][maxk], K, b;

int num[maxn];
int sum(int x) {
	int cnt = 0, tx = x;
	while(x) num[++cnt] = x % b, x /= b;
	int ans = 0, Cnt = 0;
	for(int i = cnt; i; i--) {
		for(int j = 0; j < min(2, num[i]); j++) ans += f[i][j][K-Cnt];
		if(num[i] > 1){ Cnt = -1; break; }
		Cnt += num[i];
	}
	ans += (Cnt == K);
	return ans;
}

int main() {
	f[1][0][0] = 1; f[1][1][1] = 1;
	for(int i = 1; i <= 32; i++)
		for(int j = 0; j <= 1; j++)
			for(int k = 0; k <= i; k++)
				for(int x = 0; x <= 1; x++)
					f[i+1][x][k+x] += f[i][j][k];
	
	int l = read(), r = read(); K = read(); b = read();
	printf("%d
", sum(r) - sum(l - 1));
	
	return 0;
}