GCD XOR, ACM/ICPC Dhaka 2013, UVa12716

不同的枚举方法，效率完全不同。值得记录一下！ 

#include <cstdio>
#include <cstring>
int t, a, b, c, n, cas = 0, count = 0;
int cnt[30000000];
void pre() {
    count = 0;
    memset(cnt, 0, sizeof(cnt));

    for (a = 1; a <= 30000000; a++) {
        for (c = 1; c < a; c++) {
            if (a%c == 0 && ((a-c)^a)==c) {
                count++;
            }
        }
        cnt[a-1] = count;
    }

}
int main(void)
{
    
    for (scanf("%d", &t); t--;) {
        scanf("%d", &n);
        printf("Case %d: %d
", ++cas, cnt[n-1]);
    }
    return 0;
}

#include <cstdio>
#include <cstring>
int t, a, b, c, n, cas = 0, count = 0;
const int maxn = 30000000;
int cnt[maxn];
void pre() {
    count = 0;
    memset(cnt, 0, sizeof(cnt));

    for (c = 1; c <= maxn/2; c++) {
        for (a = c+c; a <= maxn; a+=c) {
            if (((a-c)^a)==c) {
                cnt[a]++;
            }
        }
    }
    for (int i = 1; i <= maxn; i++) {
        cnt[i] += cnt[i-1];
    }

}
int main(void)
{
    pre();
    for (scanf("%d", &t); t--;) {
        scanf("%d", &n);
        printf("Case %d: %d
", ++cas, cnt[n]);
    }
    return 0;
}