POJ 2773 Happy 2006 (分解质因数＋容斥＋二分 或 欧几里德算法应用)

#include <cstdio>
#include <cstring>
#define ll long long
int const MAX = 1e6 + 6;
int fac[MAX];
int n, k, cnt;
ll sum;

void get_Factor(int x)
{
    cnt = 0;
    for(int i = 2; i <= x; i++)
    {
        if(x % i == 0)
            fac[cnt ++] = i;
        while(x % i == 0)
            x /= i;
    }
    if(x > 1)
        fac[cnt ++] = x;
}

void DFS(int pos, int num, ll cur, ll x)
{
    if(pos == cnt)
    {
        if(cur == 1)
            return;
        if(num & 1)
            sum += x / cur;
        else
            sum -= x / cur;
        return;
    }
    DFS(pos + 1, num, cur, x);
    DFS(pos + 1, num + 1, cur * fac[pos], x);
    return;
}

ll cal(ll mid)
{
    sum = 0;
    DFS(0, 0, 1, mid);
    return sum;
}

int main()
{
    while(scanf("%d %d", &n, &k) != EOF)
    {
        get_Factor(n);
        ll l = 0, r = 1e17, mid;
        while(l <= r)
        {
            mid = (l + r) >> 1;
            if(mid - cal(mid) < k)
                l = mid + 1;
            else 
                r = mid - 1;
        }
        printf("%lld\n", l);
    }
}

#include <cstdio>
int const MAX = 1e6 + 5;
int a[MAX];

int gcd(int a, int b)
{
    while(b)
    {
        int tmp = a;
        a = b;
        b = tmp % b;
    }
    return a;
}

int main()
{
    int m, k;
    while(scanf("%d %d", &m, &k) != EOF)
    {
        if(k == 1)
        {
            printf("1\n");
            continue;
        }
        if(m == 1)
        {
            printf("%d\n", k);
            continue;
        }
        int cnt = 1;
        for(int i = 1; i <= m; i++)
            if(gcd(i, m) == 1)
                a[cnt ++] = i;
        cnt --; 
        if(k % cnt == 0)
            printf("%d\n", (k / cnt - 1) * m + a[cnt]);
        else
            printf("%d\n", (k / cnt) * m + a[k % cnt]);
    }
}