HDU 2204 Eddy's爱好（容斥原理）

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 typedef long long INT;
 8 INT prim[70];
 9 
10 int dabiao()
11 {
12     int f = 0;
13     for(int i = 2;i < 60;++i)
14     {
15         int flag = 1;
16 
17         for(int j = 2;j < i;++j)
18         if(i % j == 0)
19         {
20             flag = 0;
21             break;
22         }
23         if(flag) prim[f++] = i;
24     }
25     return f;
26 }
27 
28 int calc(INT a,INT b,INT n)
29 {
30     INT s = 1;
31     while(b--)
32     {
33         if((INT)(n / a) <= s) return 0;
34         s *= a;
35     }
36     return 1;
37 }
38 int main()
39 {
40     INT n;
41     int num = dabiao();
42     while(scanf("%I64d",&n)!=EOF)
43     {
44         INT tot = 1;
45         for(int i = 0;i < num;++i)
46         {
47             INT temp = (INT)pow((double)n,1.0 / prim[i]);
48             if(temp == 1 ) break;   //开方之后不足1的话，则退出
49             tot += temp - 1;   //减1是因为每次都会统计到1
50         }
51         for(int i = 0;i < num;++i)
52         for(int j = i+1;j < num;++j)
53         {
54             INT temp = pow((double)n,1.0/(prim[i] * prim[j]));
55             if(temp == 1) break;
56             tot -= temp - 1;
57         }
58         for(int i = 0;i < num;++i)
59         for(int j = i + 1;j < num;++j)
60         for(int k = j + 1;k < num;++k)
61         {
62             INT temp = pow((double)n,1.0/(prim[i] * prim[j] * prim[k]));
63             if(temp == 1) break;
64             tot += temp - 1;
65         }
66         printf("%I64d\n",tot);
67     }
68     return 0;
69 }