hdu 1286 找新朋友 （容斥原理 || 欧拉函数）

Problem - 1286

用容斥原理做的代码：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <vector>
 6 
 7 using namespace std;
 8 
 9 const int N = 33333;
10 
11 int last[N];
12 void pre() {
13     last[1] = 1;
14     for (int i = 2; i < N; i++) {
15         if (!last[i]) {
16             for (int j = i; j < N; j += i) {
17                 last[j] = i;
18             }
19         }
20     }
21 //    for (int i = 0; i < 20; i++) cout << last[i] << endl;
22 }
23 
24 vector<int> fac;
25 void getFac(int n) {
26     fac.clear();
27     while (n > 1) {
28         fac.push_back(last[n]);
29         n /= last[n];
30     }
31     int t = (int) (unique(fac.begin(), fac.end()) - fac.begin());
32     while (fac.size() > t) fac.pop_back();
33 }
34 
35 int cntBit(int n) { return n > 0 ? cntBit(n >> 1) + (n & 1) : 0;}
36 
37 int main() {
38     pre();
39     int T, n;
40     cin >> T;
41     while (T-- && cin >> n) {
42         getFac(n);
43         int ans = 0;
44         for (int i = 1, szi = 1 << fac.size(); i < szi; i++) {
45             int tmp = 1;
46             for (int j = 0, szj = fac.size(); j < szj; j++) {
47                 if (i & 1 << j) tmp *= fac[j];
48             }
49             if (cntBit(i) & 1) {
50                 ans += n / tmp;
51             } else {
52                 ans -= n / tmp;
53             }
54         }
55         cout << n - ans << endl;
56     }
57     return 0;
58 }

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用欧拉函数做的代码：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <vector>
 6 
 7 using namespace std;
 8 
 9 const int N = 33333;
10 
11 int last[N];
12 void pre() {
13     last[1] = 1;
14     for (int i = 2; i < N; i++) {
15         if (!last[i]) {
16             for (int j = i; j < N; j += i) {
17                 last[j] = i;
18             }
19         }
20     }
21 //    for (int i = 0; i < 20; i++) cout << last[i] << endl;
22 }
23 
24 int phi(int n) {
25     int ret = 1;
26     while (n > 1) {
27         int tmp = last[n];
28 //        cout << tmp << endl;
29         ret *= tmp - 1;
30         n /= last[n];
31         while (tmp == last[n]) {
32             ret *= tmp;
33             n /= last[n];
34         }
35     }
36     return ret;
37 }
38 
39 int main() {
40     pre();
41     int T, n;
42     cin >> T;
43     while (T-- && cin >> n) cout << phi(n) << endl;
44     return 0;
45 }

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——written by Lyon

转载于:https://www.cnblogs.com/LyonLys/p/hdu_1286_Lyon.html