UVa 12716 GCD XOR

题目

题目大意

输入整数\(n\)(\(1 ≤ n ≤ 30000000\)), 有多少对整数\((a, b)\)满足: \(1 ≤ b ≤ a ≤ n\), 且\((a, b) = a ⊕ b\)(这里\((a, b)\)表示\(a\)与\(b\)的最大公约数, \(⊕\)表示按位异或运算)。例如\(n = 7\)时, 有\(4\)对: \((3, 2)\), \((5, 4)\), \((6, 4)\), \((7, 6)\)。

题解

由于最大公约数和按位异或看起来毫不相关, 可以先写个暴力程序打一些满足\((a, b) = a ⊕ b = c\)的表, 然后观察可以发现: \(c = a - b\)

证明如下: 刘汝佳说不难发现\(a - b ≤ a ⊕ b\), 且\(a - b ≥ c\)。假设存在\(c\)使得\(a - b > c\), 则\(c < a - b ≤ a ⊕ b\), 与\(c = a ⊕ b\)矛盾。

有了这个结论, 枚举\(a\)和\(c\), 计算\(b = a - c\), 则\((a, b) = (a, a - c) = c\), 此时就只需要验证\(c = a ⊕ (a - c)\)了。

代码

#include <cstdio>
int form[30000010], T, n;
int main(int argc, char **argv) {
  for (register int c(1); c <= 15000000; ++c) {
    for (register int a(c << 1); a <= 30000000; a += c) {
      if ((a ^ (a - c)) == c) ++form[a];
    }
  }
  for (register int i(2); i <= 30000000; ++i) form[i] += form[i - 1];
  scanf("%d", &T);
  for (register int i(1); i <= T; ++i) {
    scanf("%d", &n);
    printf("Case %d: %d\n", i, form[n]);
  }
  return 0;
}