【topic】
Niu Niu got a positive integer pair (x, y) from the teacher before, Niu Niu forgot how many they were.
But Niu Niu remembers that the teacher told him that neither x nor y is greater than n, and the remainder of dividing x by y is greater than or equal to k.
Niu Niu wants you to help him calculate how many possible pairs there are.
Enter description:
The input consists of two positive integers n,k (1 <= n <= 10^5, 0 <= k <= n - 1).
Output description:
For each test case, output a positive integer representing the number of possible pairs.
Example 1
enter
5 2
output
7
illustrate
The number pairs that satisfy the condition are (2,3),(2,4),(2,5),(3,4),(3,5),(4,5),(5,3)
【solve】
① https://www.nowcoder.com/questionTerminal/bac5a2372e204b2ab04cc437db76dc4f
x can be taken on [1, n], but y can only be taken on [k, n] , because there is no remainder greater than or equal to k below k.
So traverse y, for each y, count the number of matching x and add it to count.
Assuming that x can take values from [0, n], then this interval can be divided into at least (n/k) complete intervals of length y .
x = 【0，1……y-2，y-1】【y，y+1，……，2y-2，2y-1】……【……】……【……，n】
In each small interval a, the remainder of the ith number a[i]%y is i . In this way , there are yk x greater than or equal to k in each small segment (obviously, when k=0, the number of y satisfies the meaning of the question) .
【0，1，……，k，k+1，……，y-1】
In this way , the total number of traversed is (n/y)*y, and the total number of x where the condition is satisfied is (n/y)*(yk)
因为n = (n/y)*y + n%y
So there are n+1-(n/y)*y = n+1 - (n - n%y) = 1+n%y numbers that have not been traversed, let it be t.
Since n%y∈[0, y-1], then t∈[1, y].
That is to say, in my method, at least one number is left, and at most an entire interval (the number is y) is left .
But in any case, the ith number last[i]%y of this interval last must be i. Then the remainder of the last number (n) is n%y.
As a result, the interval from [k,n%y] contains a total of n%y-k+1 numbers. However, if you calculate a number less than 0, you don't need to subtract it back, just treat it as nothing.
So the last interval contains max( n %y-k+1, 0 ) x that satisfy the condition.
Finally, note that all the numbers of x∈ [0, n] are actually enumerated here. When k==0, an extra 0 is counted and must be subtracted .
// Assuming n=10, k=3, it can only be 4, 5, 6, 7, 8, 9, 10 for y
// When y=4, (n/y)*(yk) represents x When less than or equal to 8 (8 is an integer multiple of 4), there are (3,4), (7,4), Math.max(0,n%y-k+1) represents the logarithm that meets the meaning of the question when x is greater than 8 0
// When y=5, (n/y)*(yk) means x is less than or equal to 10 (10 is an integer multiple of 5), there are (3,5),(4,5),(8,5) ,(9,5), Math.max(0,n%y-k+1) means that when x is greater than 10, the logarithm that meets the meaning of the question is 0
// When y=6,(n/y)*(yk) Represents (3,6), (4,6), (5,6) when x is less than or equal to 6, Math.max(0,n%y-k+1) represents the logarithm that meets the meaning of the question when x is greater than 6 is 2, respectively (9,6), (10,6)
// When y=7, (n/y)*(yk) means x is less than or equal to 7, there are (3,7), (4,7) ,(5,7),(6,7),Math.max(0,n%y-k+1) represents that when x is greater than 7, the logarithm that meets the meaning of the question is 1, which is (10,7)
// . ..and so on
import java.util. *;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextInt();
long k = sc.nextInt();
long res = 0;
if (k == 0){
res = n * n;//The remainder of any two logarithms is greater than or equal to 0
}else {
for (long y = k + 1;y <= n;y++){
res + = (n / y) * (y - k) + Math.max(n % y - k + 1,0);
}
}
System.out.println(res);
}
}