BZOJ2045 双亲数

2301的弱化版。。。（弱过头了的说）

真是。。。为什么2301都1A了这道题却1RE+1A啊。。。蒟蒻到底了。。。

什么时候搞懂了再写题解什么的。。。

 

 1 /**************************************************************
 2     Problem: 2045
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:360 ms
 7     Memory:9596 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 = 1000005;
17 int u[N], p[N], tot;
18 bool v[N];
19  
20 inline int read(){
21     int x = 0;
22     char ch = getchar();
23     while (ch < '0' || ch > '9')
24         ch = getchar();
25          
26     while (ch >= '0' && ch <= '9'){
27         x = x * 10 + ch - '0';
28         ch = getchar();
29     }
30     return x;
31 }
32  
33 void pre_work(){
34     u[1] = 1;
35     int i, j, K;
36     for (i = 2; i < N; ++i){
37         if (!v[i])
38             p[++tot] = i, u[i] = -1;
39         for (j = 1; i * p[j] < N && j <= tot; ++j){
40             v[K = i * p[j]] = 1;
41             if (i % p[j] == 0){
42                 u[K] = 0; 
43                 break;
44             }else u[K] = -u[i];
45         }
46     }
47     for (i = 2; i < N; ++i)
48         u[i] += u[i - 1];
49 }
50  
51 ll work(int n, int m, int k){
52     n /= k, m /= k;
53     if (n > m) swap(n, m);
54     ll res = 0;
55     int i, last;
56     for (i = 1; i <= n; i = last + 1){
57         last = min(n / (n / i), m / (m / i));
58         res += (ll) (u[last] - u[i - 1]) * (n / i) * (m / i);
59     }
60     return res;
61 }
62  
63 int main(){
64     pre_work();
65     int a = read(), b = read(), k = read();
66     printf("%lld\n", work(a, b, k));
67     return 0;
68 }
View Code


 

posted on 2014-10-24 22:36  Xs酱~  阅读(237)  评论(0)  编辑  收藏  举报