POJ 2773 Happy 2006

二分 + 容斥原理

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int MAXN = 1000005;const int inf  = 0x3f3f3f3f;typedef long long ll;bool is_prime[MAXN];ll   prime[MAXN];ll   factor[110], ans;int  len, pcnt;inline ll min(ll a, ll b) {    return a < b ? a : b;}inline ll max(ll a, ll b) {    return a > b ? a : b;}void Init() {    int i, j;    len = 0;    memset(is_prime, true, sizeof(is_prime));    is_prime[0] = is_prime[1] = false;    prime[len++] = 2;    for(i = 4; i < MAXN; i += 2)        is_prime[i] = false;    for(i = 3; i * i <= MAXN; i += 2) {        if(is_prime[i]) {            prime[len++] = i;            for(j = i * i; j < MAXN; j += i)                is_prime[j] = false;        }    }    for( ; i < MAXN; i += 2)        if(is_prime[i])            prime[len++] = i;    return ;}void cal_factor(ll n) {    pcnt = 0;    for(int i = 0; i < len && prime[i] * prime[i] <= n; ++i) {        if(n % prime[i] == 0) {            factor[pcnt++] = prime[i];            while(n % prime[i] == 0)                n /= prime[i];        }    }    if(n > 1) factor[pcnt++] = n;    return ;}void dfs(int id, ll cur, int cnt, ll limit) {    if(id == pcnt) {        if(cnt & 1)            ans -= limit / cur;        else            ans += limit / cur;        return ;    }    dfs(id + 1, cur, cnt, limit);    dfs(id + 1, cur * factor[id], cnt + 1, limit);    return ;}ll get_num(ll limit) {    ans = 0;    dfs(0, 1, 0, limit);    return ans;}ll solve(ll m, ll k) {    ll ret = inf;    cal_factor(m);    int low = 1, high = inf, mid;    while(low <= high) {        mid = (low + high) / 2;        if(get_num(mid) >= k) {            ret = min(ret, mid);            high = mid - 1;        }        else {            low = mid + 1;        }    }    return ret;}int main() {    //freopen("aa.in", "r", stdin);    ll m, k;    Init();    while(cin >> m >> k) {        cout << solve(m, k) << endl;    }    return 0;}