Given X, Y, K, and B, it is required to find the number of positive integers in the range [x, y] so that they satisfy the power of k different B.
For example
X= 15,
Y= 20,
K= 2,
B= 217 = 24 + 20,
18 = 24 + 21,
20 = 24 + 22. there are three positive integers that meet the requirements to convert the problem, set g (x) to the number of <= x positive integers that meet the meaning of the question, then the difference between the original question = g (y) -G (x-1) Now solves the problem after conversion and expands X in B-base, the goal is to get a B-base string so that it only contains 0 and 1, if the value is less than or equal to X, dynamic planning can be used to resolve the problem by using F [I] [J] [flag] to perform the I-bit operation, with J numbers 1, flag = indicates the total number of satisfied requirements when the value is smaller than or equal to the value. The recurrence formula is easy to obtain, from high to low. Pay attention to the details. The final answer is f [1] [k] [0] + F [1] [k] [1] (here 1 is a percentile bit)
1 # include <iostream> 2 # include <cstring> 3 using namespace STD; 4 int X, Y, K, B; 5 Int Len, bbas [51]; 6 int f [51] [21] [2]; // 1 equals 0 less than 7 int solve (INT p) {8 Len = 0; memset (bbas, 0, sizeof (bbas); 9 memset (F, 0, sizeof (f); 10 int T = P; 11 while (t) {12 bbas [++ Len] = T % B; t = T/B; 13} 14 if (bbas [Len] = 1) {f [Len] [1] [1] = 1; F [Len] [0] [0] = 1 ;} 15 else {f [Len] [1] [0] = 1; F [Len] [0] [0] = 1;} 16 for (INT I = len-1; i> = 1; I --) 17 for (Int J = 0; j <= K; j ++) {18 if (j = 0) {19 F [I] [J] [0] = 1; 20} else {21 if (bbas [I]> 1) {22 f [I] [J] [0] = f [I + 1] [J-1] [1] + F [I + 1] [J-1] [0] + F [I + 1] [J] [1] + F [I + 1] [J] [0]; 23} else {24 if (bbas [I] = 1) {25 f [I] [J] [0] = f [I + 1] [J] [1] + F [I + 1] [J] [0] + F [I + 1] [J-1] [0]; 26 F [I] [J] [1] = f [I + 1] [J-1] [1]; 27} 28 If (bbas [I] = 0) {29 F [I] [J] [0] = f [I + 1] [J] [0] + F [I + 1] [J-1] [0]; 30 f [I] [J] [1] = f [I + 1] [J] [1]; 31} 32} 33} 34} 35 return f [1] [k] [0] + F [1] [k] [1]; 36} 37 int main () {38 CIN> x> Y> K> B; 39 cout <solve (y)-solve (x-1) <Endl; 40 return 0; 41}
Ural1057 amount of degrees