[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.
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\n", sum(r) - sum(l - 1)); return 0; }