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 = 33333;

int last[N];
void pre() {
    last[1] = 1;
    for (int i = 2; 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 > 1) {
        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 > 0 ? cntBit(n >> 1) + (n & 1) : 0;}

int main() {
    pre();
    int T, n;
    cin >> T;
    while (T-- && cin >> n) {
        getFac(n);
        int ans = 0;
        for (int i = 1, szi = 1 << fac.size(); i < szi; i++) {
            int tmp = 1;
            for (int j = 0, szj = fac.size(); j < szj; j++) {
                if (i & 1 << j) tmp *= fac[j];
            }
            if (cntBit(i) & 1) {
                ans += n / tmp;
            } else {
                ans -= n / tmp;
            }
        }
        cout << n - ans << endl;
    }
    return 0;
}

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 = 33333;

int last[N];
void pre() {
    last[1] = 1;
    for (int i = 2; 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 = 1;
    while (n > 1) {
        int tmp = last[n];
//        cout << tmp << endl;
        ret *= tmp - 1;
        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 0;
}

——written by Lyon