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：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <vector>

 using namespace std;

 const int N = ;

 int last[N];
 void pre() {
     last[] = ;
     for (int i = ; i < N; i++) {
         if (!last[i]) {
             for (int j = i; j < N; j += i) {
                 last[j] = i;
             }
         }
     }
 //    for (int i = 0; i < 20; i++) cout << last[i] << endl;
 }

 vector<int> fac;
 void getFac(int n) {
     fac.clear();
     while (n > ) {
         fac.push_back(last[n]);
         n /= last[n];
     }
     int t = (int) (unique(fac.begin(), fac.end()) - fac.begin());
     while (fac.size() > t) fac.pop_back();
 }

 int cntBit(int n) { return n >  ? cntBit(n >> ) + (n & ) : ;}

 int main() {
     pre();
     int T, n;
     cin >> T;
     while (T-- && cin >> n) {
         getFac(n);
         int ans = ;
         for (int i = , szi =  << fac.size(); i < szi; i++) {
             int tmp = ;
             for (int j = , szj = fac.size(); j < szj; j++) {
                 if (i &  << j) tmp *= fac[j];
             }
             if (cntBit(i) & ) {
                 ans += n / tmp;
             } else {
                 ans -= n / tmp;
             }
         }
         cout << n - ans << endl;
     }
     return ;
 }

用欧拉函数做的代码：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <vector>

 using namespace std;

 const int N = ;

 int last[N];
 void pre() {
     last[] = ;
     for (int i = ; i < N; i++) {
         if (!last[i]) {
             for (int j = i; j < N; j += i) {
                 last[j] = i;
             }
         }
     }
 //    for (int i = 0; i < 20; i++) cout << last[i] << endl;
 }

 int phi(int n) {
     int ret = ;
     while (n > ) {
         int tmp = last[n];
 //        cout << tmp << endl;
         ret *= tmp - ;
         n /= last[n];
         while (tmp == last[n]) {
             ret *= tmp;
             n /= last[n];
         }
     }
     return ret;
 }

 int main() {
     pre();
     int T, n;
     cin >> T;
     while (T-- && cin >> n) cout << phi(n) << endl;
     return ;
 }

——written by Lyon