（二分+容斥原理）
题意：求第$k(1\le k\le 100000000)$个与$m(1\le m\le 1000000)$互质的数字是多少？

思路：二分$k$，每次找出小于$k$且与$m$互质的数字有多少个，多了就把$k$变小，少了就把$k$放大。这里顺便复习下容斥原理的两种写法：迭代法和递归法。（一定一定注意哪些地方有可能爆int！）

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long

using namespace std;
const double eps = 1e-8;

bool vis[100100];
LL fac[100100];

/*LL solve(LL x, LL cnt){
    LL ret = 0;
    for(LL i=1; i<(1<<cnt); i++){
        LL bits = 0, mul = 1;
        for(LL j=0; j<cnt; j++)if(i&(1<<j)){
            bits ++; mul *= fac[j];
        }
        if(bits&1) ret += x/mul;
        else ret -= x/mul;
    }
    return x - ret;
}*/

LL solve(LL m, LL x, LL cnt){
    LL ret = 0;
    for(int i=x; i<cnt; i++){
        ret += m/fac[i] - solve(m/fac[i], i+1, cnt);
    }
    return ret;
}

int main()
{
    //freopen("test.txt","r",stdin);
    LL m, k;
    while(scanf("%lld%lld",&m,&k) == 2){
        LL cnt = 0, tm = m;
        for(LL i=2; i*i<=tm; i++) if(tm%i == 0){
            fac[cnt++] = i;
            while(tm%i == 0) tm /= i;
        }
        if(tm > 1) fac[cnt++] = tm;
        LL l = 1, r = 1000000000000;
        while(l < r){
            LL mid = (r-l)/2 + l;
            //if(solve(mid, cnt) >= k) r = mid;
            if(mid - solve(mid, 0, cnt) >= k) r = mid;
            else l = mid + 1;
        }
        printf("%lld\n",l);
    }
    return 0;
}