HDU 1695 GCD（莫比乌斯反演）

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 typedef long long LL;
 6 const int MAXN = 100005;
 7 
 8 bool check[MAXN];
 9 int prime[MAXN];
10 int mu[MAXN];
11 
12 void moblus()
13 {
14     memset(check, false, sizeof(check));
15     mu[1] = 1;
16     int tot = 0;
17     for (int i = 2; i < MAXN; ++i) {
18         if (!check[i]) {
19             prime[tot++] = i;
20             mu[i] = -1;
21         }
22         for (int j = 0; j < tot; ++j) {
23             if (i * prime[j] > MAXN) break;
24             check[i * prime[j]] = true;
25             if (i % prime[j] == 0) {
26                 mu[i * prime[j]] = 0;
27                 break;
28             } else {
29                 mu[i * prime[j]] = -mu[i];
30             }
31         }
32     }
33 }
34 
35 int main()
36 {
37     moblus();
38     int nCase;
39     scanf("%d", &nCase);
40     for (int cas = 1; cas <= nCase; ++cas) {
41         int a, b, c, d, k;
42         scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
43         if (k == 0) {
44             printf("Case %d: 0\n", cas);
45             continue;
46         }
47         b /= k;
48         d /= k;
49         if (b > d) swap(b, d);
50         LL ans = 0;
51         for (int i = 1; i <= b; ++i)
52             ans += (LL)mu[i] * (b / i) * (d / i);
53         LL tmp = 0;
54         for (int i = 1; i <= b; ++i)
55             tmp += (LL)mu[i] * (b / i) * (b / i);
56         ans -= tmp / 2;
57         printf("Case %d: %lld\n", cas, ans);
58     }
59     return 0;
60 }