什么东西。。。

搞了半天Mobius反演到底是什么还是没搞定。。。(至少会求了嘛。。。好不好)

但是程序写出来了^_^,可惜意义不明T T

 

 1 /**************************************************************
 2     Problem: 2301
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:10604 ms
 7     Memory:1244 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13  
14 using namespace std;
15 typedef long long ll;
16 const int N = 50005;
17 int T;
18 int u[N], p[N], tot;
19 bool v[N];
20  
21 inline int read(){
22     int x = 0;
23     char ch = getchar();
24     while (ch < '0' || ch > '9')
25         ch = getchar();
26          
27     while (ch >= '0' && ch <= '9'){
28         x = x * 10 + ch - '0';
29         ch = getchar();
30     }
31     return x;
32 }
33  
34 void pre_work(){
35     u[1] = 1;
36     int i, j, K;
37     for (i = 2; i < N; ++i){
38         if (!v[i])
39             p[++tot] = i, u[i] = -1;
40         for (j = 1; i * p[j] < N && j <= tot; ++j){
41             v[K = i * p[j]] = 1;
42             if (i % p[j] == 0){
43                 u[K] = 0; 
44                 break;
45             }else u[K] = -u[i];
46         }
47     }
48     for (i = 2; i < N; ++i)
49         u[i] += u[i - 1];
50 }
51  
52 ll work(int n, int m, int k){
53     n /= k, m /= k;
54     if (n > m) swap(n, m);
55     ll res = 0;
56     int i, last;
57     for (i = 1; i <= n; i = last + 1){
58         last = min(n / (n / i), m / (m / i));
59         res += (ll) (u[last] - u[i - 1]) * (n / i) * (m / i);
60     }
61     return res;
62 }
63  
64 int main(){
65     T = read();
66     pre_work();
67     int a, b, c, d, k;
68     ll ans;
69     while (T--){
70         a = read(), b = read(), c = read(), d = read(), k = read();
71         ans = work(b, d, k) - work(a - 1, d, k) - work(b, c - 1, k) + work(a - 1, c - 1, k);
72         printf("%lld\n", ans);
73     }
74     return 0;
75 }
View Code

 

posted on 2014-10-24 22:27  Xs酱~  阅读(202)  评论(0编辑  收藏  举报