题解 | #数对#
数对
https://www.nowcoder.com/practice/bac5a2372e204b2ab04cc437db76dc4f
#include <stdio.h> int main() { long long n = 0; long long k = 0; long long count = 0; while (scanf("%lld %lld", &n, &k) != EOF) { // 注意 while 处理多个 case if (k == 0) { printf("%lld", n * n); } else { for (long i = k + 1; i <= n; i++) { count += (n / i) * (i - k); if (n % i >= k) { count += (n % i) - k + 1; } } printf("%lld", count); } } return 0; }